Important Questions

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 10 Computer in Accounting Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 10 Computer in Accounting

1. Objective questions:

A. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
Computer, disk drives, monitors, printers, etc. are included in ____________
(a) Software
(b) Hardware
(c) Packages
(d) Personel
Answer:
(b) Hardware

Question 2.
____________ is the set of programme that direct the computer to perform the desired task.
(a) Software
(b) Hardware
(c) Packages
(d) Personel
Answer:
(a) Software

Question 3.
____________ are critical to the success of any endeavor because people operate the system.
(a) Software
(b) Hardware
(c) Packages
(d) Personel
Answer:
(d) Personel

Question 4.
Customized software is prepared to meet ____________ of the customer.
(a) special requirement
(b) easiness
(c) savings
(d) basic need
Answer:
(a) special requirement

Question 5.
Tailored Software is prepared for the ____________ size business.
(a) small
(b) medium
(c) large
(d) small and medium
Answer:
(c) large

Question 6.
____________ application can be downloaded and install from the websites.
(a) Ready to use
(b) Customized
(c) Tailored
(d) Free and open source
Answer:
(d) Free and open source

B. Give the word term or phrase which can substitute each of the following statements:

Question 1.
The software which is useful for small scale firms/business.
Answer:
Ready to use Software

Question 2.
Software prepared for multi-users and geographically scattered locations.
Answer:
Tailored Software

Question 3.
Details of bank transactions are maintained in this book.
Answer:
Bank book

Question 4.
The software is prepared for business transaction automation.
Answer:
Accounting Software

Question 5.
Software used for demo purposes with all major features.
Answer:
Demo Software

Question 6.
Fully functional, safe, and legal software.
Answer:
Legal Software

C. State whether the following statements are true or false with reason:

Question 1.
In Tally, the F7 key is for credit purchase or sales transactions.
Answer:
This statement is False.
In Tally F7 function key is for journal vouchers and it should not be used for credit purchase or sales transactions.

Question 2.
Pirated software is fully functional software.
Answer:
This statement is False.
Pirated software is nearly fully functional software but its use is illegal and risky.

Question 3.
Computerized Accounting is better than manual accounting.
Answer:
This statement is True.
Due to computerized Accounting, firms or businesses can save time, expenses and get the best output within no time which is not possible in manual accounting therefore computerized accounting is better than manual accounting.

Question 4.
In the computerized accounting system, closing balances are carried forward to the next period, automatically.
Answer:
This statement is True.
Just like a manual accounting system, closing balances are not to be carried forward for the next period, in a computerized accounting system. As per the program it is automatically carried forward.

D. Answer in One Sentences:

Question 1.
Write the full form of CAS, MIS, and GAPP.
Answer:
Full form of:
CAS – Computerized Accounting System.
MIS – Management Information System.
GAAP – Generally Accepted Accounting Principles.

Question 2.
Write the components which form Computerized Accounting System.
Answer:
Hardware, Software, and company Personnel are the components that form Computerized Recounting System.

Question 3.
Write the names of equipment that are included in the Hardware.
Answer:
Electronic equipment like computers, hard disks, monitors, printers, and the network that connects with them, etc. are included in Hardware.

Question 4.
In accounting software, generally which components are used?
Answer:
In accounting software generally, the following components are used:

1. Creation of Accounting Documents
2. Recording of Transaction.
3. Preparation of Trial Balance and Financial Statements.

Question 5.
State the names of Accounting Packages.
Answer:
Ready to use, Customized, Tailored and Free, and open source are the different names of Accounting Packages.

Question 6.
State the names of Accounting Software.
Answer:
Legal/Licensed software, Demo software, and Pirated software are different accounting software.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 9 Analysis of Financial Statements Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 9 Analysis of Financial Statements

Objective Questions

A. Select the most appropriate alternative from those given below and rewrite the sentences:

Question 1.
The methodical classification of financial statement is called ____________
(a) an interpretation
(b) an analysis
(c) ratio
(d) Profit and Loss A/c
Answer:
(b) an analysis

Question 2.
The short-term deposits are ____________
(a) net cash
(b) cash equivalent
(c) cashflow
(d) cash outflow
Answer:
(b) cash equivalent

Question 3.
Cash proceeds from issue of debentures is a ____________ activity.
(a) financial
(b) non-financial
(c) operating
(d) trading
Answer:
(a) financial

Question 4.
The relationship between net profit before tax, interest and dividend and capital employed is known from ____________
(a) Current ratio
(b) Quick ratio
(c) ROI
(d) ROCE
Answer:
(c) ROI

Question 5.
Bills receivable is ____________
(a) Liquid asset
(b) Net profit
(c) Current asset
(d) Net loss
Answer:
(c) Current asset

Question 6.
ROCE should be ____________ than ROI.
(a) less
(b) higher
(c) equal
(d) none of these
Answer:
(b) higher

Question 7.
Ideally liquid ratio/quick ratio should be ____________
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 1 : 3
Answer:
(b) 1 : 1

Question 8.
Gross/Net profit ratio expressed in ____________
(a) number
(b) ratio
(c) percentage
(d) words
Answer:
(c) percentage

B. Give one word/term/phrase for each of the following statements.

Question 1.
The tool for analysis of financial statement where individual figures of Balance Sheet are converted into a percentage.
Answer:
Common Size Balance Sheet

Question 2.
The type of activity in cash flow analysis, involving the purchase of fixed assets.
Answer:
Investing Activity

Question 3.
The ratio measures the efficiency of the production department.
Answer:
Gross Profit Ratio

Question 4.
The ratio measures the overall efficiency of the business.
Answer:
Net Profit Ratio

Question 5.
The ratio shows the operational efficiency of the business.
Answer:
Operating Profit Ratio

Question 6.
The ratio is computed to measure the overall efficiency or profitability of the business.
Answer:
Return On Investment (ROI)

C. Answer in one sentence only.

Question 1.
Define financial statements.
Answer:
The statements which are prepared by the business to find out profitability, efficiency, solvency, growth of the business to judge the financial strength and status are called financial statements.

Question 2.
Who prepares financial statements?
Answer:
The financial statements are prepared by the Profit-making organisations as well as Non-profit concerns or organisations.

Question 3.
State the main tools or techniques of financial analysis.
Answer:
The main tools or techniques of financial analysis are as follows:

1. Comparative financial statement
2. Common size statement
3. Cash flow analysis.

Question 4.
State the primary objective of the cash flow statement.
Answer:
The primary objective of the cash flow statement is to help management in taking decisions and making a plan by providing current information on cash inflow and outflow of any particular period.

Question 5.
What is Financial Ratio?
Answer:
A financial ratio is a mathematical number that measures the relationship between two accounting figures.

Question 6.
Write the names of the Balance Sheet ratio.
Answer:
Balance Sheet ratios are

• Current ratio
• Liquid ratio.

Question 7.
Write the names of the Income Statement ratio.
Answer:
Income Statement ratios are-

• Gross Profit ratio
• Net Profit ratio
• Operating Expense ratio.

Question 8.
Give three examples of current liability.
Answer:
Examples of current liability are Sundry creditors, Bills payable, Bank overdraft, Short-term loans, etc.

Question 9.
Write the names of combined/mixed ratios.
Answer:
Names of combined/mixed ratios are-

1. Return On Capital Employed (ROCE)
2. Return On Investment (ROI)

Solved Problems

Question 1.
Calculate Current Ratio:
Debtors = ₹ 90,000, Creditors = ₹ 30,000, Bills receivables = ₹ 10,000, Bills payable = ₹ 12,000, Stock of goods = ₹ 40,000, Short-term loan = ₹ 40,000, Outstanding expenses = ₹ 14,000, Cash balance = ₹ 70,000, Machinery = ₹ 1,00,000, Current investments = ₹ 25,000, Non-current investments = ₹ 25,000, Loose tools = ₹ 15,000, Bank overdraft = ₹ 29,000.
Solution:
Current assets = Debtors + Bills receivable + Stock of goods + Cash balance + Current investments + Loose tools
= 90,000 + 10,000 + 40,000 + 70,000 + 25,000 + 15,000
= ₹ 2,50,000
Current liabilities = Creditors + Bills payable + Short-term loan + Outstanding expenses + Bank overdraft
= 30,000 + 12,000 + 40,000 + 14,000 + 29,000
= ₹ 1,25,000
Current ratio = \(\frac{\text { Current assets }}{\text { Current liabilities }}\)
= \(\frac{2,50,000}{1,25,000}\)
= 2 : 1
Note: Machinery and Non-current investment are to be committed as it is not to be included in current assets.

Question 2.
Calculate Quick Ratio:
Working capital = ₹ 1,70,000, Prepaid expenses = ₹ 10,000, Inventory/Stock = ₹ 15,000, Prepaid expenses = ₹ 10,000, Current liabilities = ₹ 1,25,000, Bank overdraft = ₹ 35,000
Solution:
Current assets = Current liabilities + Working capital
= 1,25,000+ 1,70,000
= ₹ 2,95,000
Quick assets = Current assets – Inventory – Prepaid expense
= 2,95,000 – 15,000 – 10,000
= ₹ 2,70,000
Quick liabilities = Current liabilities – Bank overdraft
= 1,25,000 – 35,000
= ₹ 90,000
Quick ratio = \(\frac{\text { Quick assets }}{\text { Quick liabilities }}\)
= \(\frac{2,70,000}{90,000}\)
= 3 : 1

3. Calculate Gross Profit Ratio:
Opening stock = ₹ 20,000, Closing stock = ₹ 25,000, Purchases = ₹ 1,00,000, Purchase return = ₹ 10,000, Sales = ₹ 2,25,000, Sales return = ₹ 15,000, Direct expenses = ₹ 20,000.
Solution:
Cost of goods sold = Opening stock + Purchases – Purchase return + Direct expenses – Closing stock
= 20,000 + 1,00,000 – 10,000 + 20,000 – 25,000
= 1,30,000 – 25,000
= ₹ 1,05,000
Net sales = Sales – Sales return
= 2,25,000 – 15,000
= ₹ 2,10,000
Gross profit = Net sales – Cost of goods sold
= 2,10,000 – 1,05,000
= ₹ 1,05,000
Gross Profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
= \(\frac{1,05,000}{2,10,000} \times 100\)
= 50%

Question 4.
Calculate Gross Profit Ratio :
Sales ₹ 9,00,000, Gross profit ratio 20% on cost.
Solution:
Gross profit is 20% on cost.
Goods costing ₹ 100 must have been sold for ₹ 120.
Hence, if sales is ₹ 120, gross profit is ₹ 20.
If sales is ₹ 9,00,000 then Gross profit =?
Gross profit = 9,00,000 × \(\frac{20}{100}\) = ₹ 1,50,000
Gross profit ratio = \(\frac{\text { Gross profit }}{\text { Net sales }} \times 100\)
= \(\frac{1,50,000}{9,00,000} \times 100\)
= 16.67%

Question 5.
Calculate Net Profit Ratio:
Sales = ₹ 10,00,000, Cost of goods sold = ₹ 4,20,000, Indirect expenses = ₹ 30,000, Administrative expenses = ₹ 1,00,000, Selling and Distribution expenses = ₹ 80,000, Interest on debentures shares = ₹ 40,000.
Solution:
Gross profit = Sales – Cost of goods sold
= 10,00,000 – 4,20,000
= ₹ 5,80,000
Net profit = Gross profit – Administrative expenses – Selling and Distribution expenses – Indirect expenses – Interest on debentures
= 5,80,000 – 1,00,000 – 80,000 – 30,000 – 40,000
= ₹ 3,30,000
Net profit ratio = \(\frac{\text { Net profit }}{\text { Sales }} \times 100\)
= \(\frac{3,30,000}{10,00,000} \times 100\)
= 33%

Question 6.
Calculate Operating Ratio:
Trading and Profit and Loss A/c of Kalpana for the year ending 31st March 2019.

Solution:
Cost of goods sold = Net sales – Gross profit
= 7,70,000 – 3,90,000
= ₹ 3,80,000
Operating expenses = Adm. exp. + Selling and Distribution expenses
= 50,000 + 60,000
= ₹ 1,10,000
Operating ratio = \(\frac{\text { Cost of goods sold }+\text { Operating expense }}{\text { Net sales }} \times 100\)
= \(\frac{3,80,000+1,10,000}{7,70,000} \times 100\)
= \(\frac{4,90,000}{7,70,000} \times 100\)
= 63.64%

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics

Question 1.
Express the speed of EM waves in terms of permittivity and permeability of the medium.
Answer:
In a material medium, the speed of EM waves is given by, c = \(\sqrt{\frac {1}{εµ}}\)
Where, ε = Permittivity and µ = permeability.
These constants depend on the electric and magnetic properties of the medium.

Question 2.
How can one classify commonly observed phenomena of light on the basis of nature of light?
Answer:
Commonly observed phenomena concerning light can be broadly split into three categories:

1. Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.
2. Wave optics or physical optics: Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
3. Particle nature of light: Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Question 3.
State the fundamental laws on which ray optics is based.
Answer:
Ray optics is based on the following fundamental laws:
i. Light travels in a straight line in a homogeneous and isotropic medium.

ii. Two or more rays can intersect at a point without affecting their paths beyond that point.

iii. Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

iv. Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (90 and angle of refraction (62) are related by Snell’s law, given by,
n₁ sin θ₁ = n₂ sin θ₂
where, n₁, n₂ = refractive indices of medium 1 and medium 2 respectively.

Question 4.
Explain Cartesian sign conventions using a graph.
Answer:
According to Cartesian sign conventions:
i. All distances are measured from the optical centre or pole.

ii. Figures should be drawn in such a way that the incident rays travel from left to right. Thus, a real object should be shown to the left and virtual object or image to the right of pole (or optical centre).

iii. X-axis can be conveniently chosen as the principal axis with origin at the pole.

iv. Distances to the left of the pole are negative and those to the right of the pole are positive.

v. Distances above the principal axis (X-axis) are positive while those below it are negative.

Question 5.
Define and represent in a neat diagram the following terms:
i. Diverging beam
ii. Converging beam
Answer:
i. A diverging beam of light corresponds to rays of light coming from real point object.

ii. A converging beam corresponds to rays of light directed to a virtual point object or image.

Question 6.
Thickness of the glass of a spectacle is 2 mm and refractive index of its glass is 1.5. Calculate time taken by light to cross this thickness. Express your answer with most convenient prefix attached to the unit ‘second’.
Answer:
Speed of light in vacuum, c = 3 × 10⁸ m/s
Given that:
Refractive index, (n_glass) = 1.5
Thickness of the glass = 2 mm
= 2 × 10sup>-3 m
∵ Re fractive index (n_glass) = \(\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in glass (v) }}\)
∵ v = \(\frac{\mathrm{c}}{\mathrm{n}_{\text {glass }}}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 10⁸ m/s
As v = \(\frac {s}{t}\)
time taken (t) to cross the thickness (s),
t = \(\frac{\mathrm{s}}{\mathrm{v}}=\frac{2 \times 10^{-3}}{2 \times 10^{8}}\) = 1 × 10^-11 s
Most convenient unit to express this small time is nano second. (1 ns = 10^-9 s)
∴ t = 0.01 × 10^-9 s = 0.01 ns

Question 7.
Explain the properties of the image formed after reflection of light from a plane surface.
Answer:

1. The image of an object kept in front of a plane reflecting surface is virtual and laterally inverted.
2. Image is of the same size as that of the object.
3. It is situated at the same distance as that of object but on the other side of the reflecting surface.

Question 8.
Explain the formula to find the number of images formed when an object is placed in between two plane mirrors inclined at an angle θ.
Answer:

1. If an object is kept between two plane mirrors inclined at an angle θ, multiple images (N) are formed due to multiple reflections from both the mirrors.
2. The number of images can be calculated using formula n = \(\frac {360}{θ}\)
3. Exact number of images seen (N) depends upon the angle between the mirrors and position of the object.
4. When n is an even integer, for all positions of the object the number of images formed are N = n – 1.
5. When n is an odd integer:
   a. For an object placed at the angle bisector of the mirrors: N = n- 1
   b. For an object placed off the angle bisector of the mirrors: N = n
6. If n is not an integer, N = m, where m is integral part of n.

Question 9.
Define radius of curvature of a spherical mirror.
Answer:
Radius of the sphere of which a mirror is a part is called as radius of curvature of the mirror.

Question 10.
What is the focal length of a spherical mirror? Give its relation with the radius of curvature.
Answer:
i. For a concave mirror focal length is the distance at which parallel incident rays converge. For a convex mirror, it is the distance from where parallel rays appear to be diverging after reflection.

ii. In case of spherical mirrors, half of radius of curvature is focal length of the mirror,
f = \(\frac {R}{2}\)

Question 11.
Show with the help of a ray diagram that focal length of convex mirror is positive while that of concave mirror is negative.
Answer:
i. According to sign conventions, the incident rays are drawn from left of the mirror to the right as shown in the ray diagrams below.

ii. As the rays incident on convex mirror appear to converge at a point on the positive side of the origin, the focal length of the convex mirror is positive.

iii. However, in case of concave mirror, the rays converge at a point on negative side of the origin, the focal length of the concave mirror is negative.

Question 12.
Give relation between focal length, object distance and image distance for a small spherical mirror.
Answer:
For a point object or for a small finite object, the focal length of a small spherical mirror is related to object distance and image distance as,
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\)

Question 13.
What is lateral magnification? How does it vary in different types of spherical mirrors?
Answer:
i. Ratio of linear size of image to that of the object, measured perpendicular to the principal axis, is defined as the lateral magnification.
∴ m = \(\frac {h_2}{h_1}\) = \(\frac {v}{u}\) (for spherical lenses)
m = –\(\frac {v}{u}\) (for spherical mirrors)

ii. For any position of the object, a convex mirror always forms virtual, erect and diminished image. Thus, lateral magnification for convex lens is always m < 1.

iii. In the case of a concave mirror, it depends upon the position of the object.

Question 14.
Complete the following table for a concave mirror?

Position of object	Position of image	Lateral magnification
U = ∞	v = f	m = 0
u > 2f	…………….	m < 1
u = f	V = ∞	………………
……………	|v| > |u|	m > 1
2f > u > f	………………	m > 1

Answer:

Position of object	Position of image	Lateral magnification
U = ∞	v = f	m = 0
u > 2f	2f > v > f	m < 1
u = f	V = ∞	m = ∞
u < f	|v| > |u|	m > 1
2f > u > f	v > 2f	m > 1

Question 15.
Explain with proper diagram why parabolic mirrors are preferred over spherical one.
Answer:
i. Unlike spherical shape, every point on a parabola is equidistant from a straight line and a point.

ii. Consider given parabola having RS as directrix and F as the focus. Points A, B, C on it are equidistant from line RS and point F.

iii. Hence A’A = AF, B’B = BF, C’C = CF, and so on.

iv. If rays of equal optical path converge at a point, that point is the location of real image corresponding to that beam of rays.

v. From figure, the paths A”AA’, B”BB’. C”CC’, etc., are equal paths when mirror is neglected.

vi. If the parabola ABC is a mirror then by definition of parabola the respective optical paths,
A”AF = B”BF = C”CF

vii. Thus, F is the single point focus for entire beam of rays parallel to the axis and there is no spherical aberration.
Hence, parabolic mirrors are preferred over spherical one as there is no spherical aberration.

Question 16.
A small object is kept symmetrically between two plane mirrors inclined at 38°. This angle is now gradually increased to 41°, the object being symmetrical all the time. Determine the number of images visible during the process.
Answer:
The object is kept symmetrically between two plane mirrors. This implies the object is placed at angle bisector.
Thus, for θ = 38°,
n = \(\frac {360}{38}\) = 9.47
As it is not integral, N = 9 (the integral part of n)
∴ For going from 38° to 41°, the mirrors go through angles 39° and 40°. Number of images formed will remain 9 for all angles between 38° and 40°.
For angles > 40°, the n goes on decreasing and when θ = 41°,
n = \(\frac {360}{41}\) = 8.78 i.e., N = 8

Question 17.
A thin pencil of length 20 cm is kept along the principal axis of a concave mirror of curvature 30 cm. Nearest end of the pencil is 20 cm from the pole of the mirror. What will be the size of image of the pencil?
Answer:
For the pencil kept along the principal axis and the end of the pencil closest to pole is at 20 cm,
say, u₁ = -20 cm
Flence, the other end of the stick is at distance, u₂ = (u₁ + 20) = -40 cm from pole of the mirror.
As R = -30 cm, F = \(\frac {R}{2}\) = -15 cm

∴ v₂ = -24 cm
Here, negative signs indicate that images are formed on the left of the mirror.

The length of the image formed is given by,
v = v₂ – v₁ = -24 – (-60) = 36 cm.

Question 18.
An object is placed at 15 cm from a convex mirror having radius of curvature 20 cm. Find the position and kind of image formed by it.
Answer:
Given: u = – 15 cm,
f = \(\frac {R}{2}\) = + \(\frac {20}{2}\) = + 10cm
To find: Nature and position of image (v)
Formula: \(\frac {1}{v}\) + \(\frac {1}{u}\) = \(\frac {1}{f}\)
Calculation:
From formula,
∴ \(\frac {1}{v}\) = \(\frac {1}{f}\) – \(\frac {1}{u}\)
= \(\frac {1}{+10}\) – \(\frac {1}{-15}\) = \(\frac {1}{10}\) + \(\frac {1}{15}\)
= \(\frac {2+3}{30}\) = \(\frac {5}{30}\) = \(\frac {1}{6}\)
∴ v = 6 cm

Question 19.
Prove that refractive index of a glass slab is given by the formula,
n = \(\frac {Real depth}{Apparent depth}\)
Answer:
i. Consider a plane parallel slab of a transparent medium of refractive index n.

ii. A point object O at real depth R appears to be at I at apparent depth A, when seen from outside (air).

iii. Consider incident ray OA and OB as shown in figure.

iv. Assuming i and r to be very small, we can write,
tan r = \(\frac {x}{A}\) ≈ sin r and tan i = \(\frac {x}{R}\) ≈ sin i

v. According to Snell’s law, for a ray travelling from denser medium to rarer medium,
n = \(\frac{\sin \mathrm{r}}{\sin \mathrm{i}} \approx \frac{\left(\frac{\mathrm{x}}{\mathrm{A}}\right)}{\left(\frac{\mathrm{x}}{\mathrm{D}}\right)}=\frac{\mathrm{R}}{\mathrm{A}}=\frac{\text { Real depth }}{\text { Apparent depth }}\)

Question 20.
The depth of a pond is 10 m. What is the apparent depth for a person looking normally to the water surface? n_water = 4/3.
Answer:
Given: Real depth of pond, d_real = 10 m,
n_w = \(\frac {4}{3}\)
To find: Apparent depth
Formula: n = \(\frac {Realdepth}{Apparent depth}\)
Calculation: From formula,
∴ Apparent depth = \(\frac {Realdepth}{n}\) = \(\frac {10}{(\frac{4}{3})}\)
= \(\frac {10×3}{4}\) = 7.5 m

Question 21.
A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water surface. How much deep should the crane immerse its beak to pick that fish?
For the fish, how much above the water surface does the crane appear? Refractive index of water = 4/3.
Answer:
For crane, apparent depth of fish = 6 cm,
Given that refractive index (n_w) = \(\frac {4}{3}\)
n_w = \(\frac {Realdepth}{Apparent depth}\)
∴ Apparent depth = \(\frac {4}{3}\) × 6 = 8 cm
Similarly, for fish, real height of crane = 6 m and
\(\frac{\mathrm{n}_{\mathrm{air}}}{\mathrm{n}_{\mathrm{w}}}=\frac{1}{\mathrm{n}_{\mathrm{w}}}=\frac{\text { Real height }}{\text { Apparent height }}\)
\(\frac {3}{4}\) = \(\frac {6}{Apparent height}\)
i.e., Apparent height = \(\frac {4×6}{3}\) = 8 m

Question 22.
Write a short note on Periscope.
Answer:
i. Instrument used to see the objects on the surface of a water body from inside of water is called periscope.
ii. It consists of two right angled prisms. The incident rays of light are reflected twice through these prisms.
iii. Total internal reflections occur inside these prisms and a clear view of the surface of water is obtained.

Question 23.
A ray of light passes from glass (n_g = 1.52) to water (n_w = 1.33). What is the critical angle of incidence?
Answer:
Given: n_g = 1.52, n_w = 1.33
To find: Critical angle (i_c)
formula: sin i_c = \(\frac {n_2}{n_1}\) = \(\frac {n_w}{n_g}\)
Calculation:
From formula,
i_c = sin^-1 (\(\frac {1.33}{1.52}\)) = sin^-1 (0.875) = 61°2′

Question 24.
There is a tiny LED bulb at the center of the bottom of a cylindrical vessel of diameter 6 cm. Height of the vessel is 4 cm. The beaker is filled completely with an optically dense liquid. The bulb is visible from any inclined position but just visible if seen along the edge of the beaker. Determine refractive index of the liquid.
Answer:

As the bulb is just visible from the edge, the angle of incidence formed by ray OP must be equal to critical angle.
∴ refractive index (n) = \(\frac {1}{sin i_c}\)
From Figure,
tan i_c = \(\frac {PQ}{OQ}\) = \(\frac {4}{3}\)
∴ sin i_c = \(\frac {OQ}{OP}\) = \(\frac {3}{5}\)
∴ n_liquid = \(\frac {5}{3}\)

Question 25.
What are convex and concave lenses? For which condition, convex lens will have negative focal length?
Answer:

1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive.
2. Convex lens is visualized to be internal cross section of two spheres (or one sphere or a plane surface).
3. A lens is concave if it is thicker at periphery and narrows down towards centre and has negative focal length.
4. Concave lens is visualized to be external cross section of two spheres.
5. For lenses of material optically rarer than the medium in which those are kept, convex lenses will have negative focal length and they will diverge the incident rays.

Question 26.
Which lenses can be considered as thin lenses?
Answer:
Lenses for which the maximum thickness is at least 50 times smaller than all the other distances are considered as thin lenses.

Question 27.
Give the expression for the focal length of combination of lenses when
i. Lenses are kept in contact with each other
ii. Two lenses kept at a distance d apart from each other.
Answer:
i. For thin lenses kept in contact:
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}+\frac{1}{\mathrm{f}_{3}}\) + ………

ii. For two lenses kept distance d apart:
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}\)

Question 28.
An object is placed infront of a convex surface separating two media of refractive index 1.1 and 1.5. The radius of curvature is 40 cm. Where is the image formed when an object is placed at 220 cm from the refracting surface?
Solution:
Given: n₁ = 1.1, n2 = 1.5, R = + 40 cm,
u = -220 cm
To find: Position of image (v)
Formula: \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\)
Calculation: From formula,

Question 29.
A glass paper-weight (n = 1.5) of radius 3 cm has a tiny air bubble trapped inside it. Closest distance of the bubble from the surface is 2 cm. Where will it appear when seen from the other end (from where it is farthest)?
Answer:
From figure, distance OR = 2 cm
∴ Distance OP = 4 cm

According to sign conventions,
OP = u = -4 cm and CP = R = -3 cm
For refraction at curved surface,

Question 30.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
Given: n = 1.55, f = 20 cm,
R₁ = R and R₂ = – R
(By sign convention)
To Find: Radius of curvature (R)
Formula: \(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula,
\(\frac {1}{20}\) = (1.55 – 1) \(\left[\frac{1}{R}-\left(-\frac{1}{R}\right)\right]=0.55 \times \frac{2}{R}\)
∴ R = \(\frac {1.10}{1}\) × 20 = 22 cm

Question 31.
A dense glass double convex lens (n = 2) designed to reduce spherical aberration has |R₁| : |R₂| = 1:5. If a point object is kept 15 cm in front of this lens, it produces its real image at 7.5 cm. Determine R₁ and R₂.
Answer:
Given: |R₁| : |R₂| = 1 : 5, u = -15 cm,
v = +7.5 cm, n = 2
To Find: Radii of curvature of double convex lens (R₁) and (R₂)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. \(\frac {1}{f}\) = (n – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula (i),
\(\frac{1}{f}=\frac{1}{7.5}-\frac{1}{(-15)}=\frac{1}{5}\)
∴ f = +5 cm
Substituting this value in formula (ii), we get,
\(\frac{1}{5}=(2-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
∴ \(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}=\frac{1}{5}\)
By sign conventions,
\(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\left(-\mathrm{R}_{2}\right)}=\frac{1}{5}\) ………….. (1)
Also \(\frac{\left|R_{1}\right|}{\left|R_{2}\right|}=\frac{1}{5}\)
∴ |R₂| = 5 |R₁| …………… (2)
Substituting in equation (1),
∴ \(\frac{1}{R_{1}}-\frac{1}{\left(-5 R_{1}\right)}=\frac{1}{5}\)
∴ \(\frac {6}{5R_1}\) = \(\frac {1}{5}\)
∴ R₁ = 6 cm
Using in equation (2),
R₂ = 5 × 6 = 30 cm

Question 32.
Why are prism preferred for dispersion over two parallel surfaces? Explain its construction in brief.
Answer:
i. In case of two parallel surfaces, for dispersion to be easily detectable, they must be separated over a large distance.
ii. In order to have appreciable and observable dispersion, two parallel surfaces are not useful. In such case we use prisms, in which two refracting surfaces inclined at an angle.
iii. Commonly used prisms have three rectangular surfaces forming a triangle.
iv. Two of which take part in refraction at a time. The one, not involved in refraction is called base of the prism.
v. Any section of prism perpendicular to the base is called principal section of the prism. Commonly all the rays considered during refraction lie in this plane.

Question 33.
Draw neat labelled diagrams showing refraction of a monochromatic light and white light through a prism.
Answer:

Question 34.
For a prism prove that i + e = A + δ where the symbols have their usual meanings.
Answer:
i. Consider a principal section ABC of a prism of absolute refractive index n kept in air as shown in figure.

ii. Let A be the refracting angle of prism and surface BC be the base.

iii. A monochromatic ray PQ obliquely strikes first reflecting surface AB such that, angle of incidence ∠PQM at Q is i.

iv. After refraction at Q, the ray deviates towards the normal and strikes second refracting surface AC at R which is the point of emergence.

v. Angles of refraction at Q (∠NQR) and at R (∠QRN) are r₁ and r₂ respectively.

vi. After R. the ray deviates away from normal and finally emerges along RS making e as the angle of emergence.

vii. Emergent ray RS meets an extended incident ray QT at X if traced backward. In this case, ∠TXS gives the angle of deviation.

viii. From figure,
∠AQN = ∠ARN = 90°
∴ From quadrilateral AQNR,
A + ∠QNR = 180° ………. (1)
From ∆ QNR,
r₁ + r₂ + ∠QNR = 180° ………. (2)
∴ A = r₁ + r₂ ……… (3)

ix. Angle δ forms an exterior angle for ∆ XQR.
∴ ∠XQR + ∠XRQ = δ
∴ (i – r₁) + (e – r₂) = δ
∴ (i + e) – (r₁ + r₂) = δ
From equation (3),
δ = i + e – A
∴ i + .e = A + δ

Question 35.
Explain δ versus i curve for refraction of light through a prism.
Answer:
i. Variation of angle of incidence i with angle of deviation δ is as shown in figure.

ii. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum (δ_m) value and then increases.

iii. The curve shown in the figure is not a symmetric parabola, but the slope in the part towards right is less.

iv. Except at δ = δ_m there are two values of i for any given δ. From principle of reversibility of light, we can conclude that if one of these values is i, the other must be e and vice versa. Thus at δ = δ_m, we have i = e.

Question 36.
Show that, at condition of minimum deviation, n = \(\frac{\sin \left(\frac{\mathbf{A}+\boldsymbol{\delta}_{\mathrm{m}}}{\mathbf{2}}\right)}{\sin \left(\frac{\mathbf{A}}{\mathbf{2}}\right)}\)

i. For every angle of deviation except angle of minimum deviation, there are two values of angle of incidence.

ii. However, at angle of minimum deviation there is only one corressponding angle of incidence.

iii. From principle of reversibility in path PQRS, the values of i and e are interchangeable for every δ. Thus, at minimum deviation, i = e.

iv. This implies the angles of refraction r₁ and r₂ are also equal. Also, A = r₁ + r₂
∴ A = 2 r i.e., r = \(\frac {A}{2}\) ……. (1)

v. In case of minimum deviation, QR is parallel to base BC and the figure is symmetric.

vi. Using i + e = A + δ,
i + i = A + δ_m
∴ i = \(\frac {A+δ_m}{2}\) …………….(2)

vii. According to Snell’s law,
n = \(\frac {sin i}{sin r}\)
Thus, using equations (1) and (2),
n = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
This is the prism formula.

Question 37.
For grazing emergence of a ray in a prism, find out minimum possible values for angle of incidence (i) and angle of refraction (r₁) for commonly used glass prism.
Answer:

i. At the point of emergence in prism, the ray travels from a denser medium into rarer.
Thus, if r₂ = sin^-1 (\(\frac {1}{n}\)) is the critical angle, the angle of emergence e = 90°. This is called grazing emergence.

ii. Angle of prism A is constant for a given prism and A = r₁ + r₂. Hence the corresponding r₁ and i will have their minimum possible values.

iii. For commonly used glass prisms,
n = 1.5 (r₂)_max = sin^-1 (\(\frac {1}{n}\)) = sin^-1 (\(\frac {1}{1.5}\)) = 41.49°

iv. If, prism is symmetric (equilateral),
A = 60°
∴ r₁ = 60° – 41°49′ = 18°11′
∴ n = 1.5 = \(\frac{\sin \left(\mathrm{i}_{\min }\right)}{\sin \left(18^{\circ} 11^{\prime}\right)}\)
sin (i_min) = 1.5 × sin (18°11′)
∴ ii_min = 27°55′ ≅ 28°.

Question 38.
Derive the formula for angle of deviation for thin prisms.
OR
Show that in a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).
Answer:
For thin prisms (refracting angle < sin 10°). sin θ ≈ θ
∴ Refractive index, n = \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}_{1}} \approx \frac{\mathrm{i}}{\mathrm{r}_{1}}\)
Also n = \(\frac{\sin e}{\sin \mathrm{r}_{2}} \approx \frac{\mathrm{e}}{\mathrm{r}_{2}}\)
∴ i ≈ n r₁ and e ≈ nr₂

ii. Substituting this in, i + e = A + δ, we get,
i + e = n (r₁ + r₂) = nA = A + δ
∴ S = A(n – 1)
A and n are constant for a given prism. Thus, for a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

Question 39.
Give the expression for mean deviation for a beam of white light.
Answer:
For a beam of white light, yellow colour is practically chosen to be the mean colour for violet and red.
This gives mean deviation as,
δ_VR = \(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\) ≈ δ_Y = A(n_Y – 1)
Where, n_Y = refractive index for yellow colour.

Question 40.
A fine beam of white light is incident upon the longer side of a plane parallel glass slab of breadth 5 cm at angle of incidence 60°. Calculate lateral deviation of red and violet rays and lateral dispersion between them as they emerge from the opposite side. Refractive indices of the glass for red and violet are 1.51 and 1.53 respectively.
Answer:
As shown in figure,
VM = L_V = lateral deviation for violet colour,
RT = L_R = lateral deviation for red colour,
∴ Lateral dispersion between these colours, L_VR = L_V – L_R

∴ r_R = sin^-1 (0.5735) ≈ 35°
Similarly,
sin r_V = \(\frac{\sin 60^{\circ}}{1.53}=\frac{\sqrt{3}}{2 \times 1.53}=\frac{1.732}{3.06}\)
= antilog {log (1.732) – log (3.06)}
= antilog {0.2385 – 0.4857}
= antilog 11.7528}
= 0.5659
∴ sin r_V = 0.566
∴ r_V = sin^-1 (0.566) = 34°28′
∴ Angle of deviation for red colour
= i – r_R = 60° – 35° = 25°
and that for violet colur = i – r_V = 60° – 34°28′
= 25°32′
From figure, in ∆ANR,
AR = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{R}}}=\frac{5}{\cos \left(35^{\circ}\right)}=\frac{5}{0.8192}\) = 6.104 cm
Similarly ∆ANV,
AV = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{V}}}=\frac{5}{\cos \left(34^{\circ} 28^{\prime}\right)}=\frac{5}{0.8244}\) = 6.065 cm
∴ For red colour, L_R = RT = AR [sin(i – r_R)]
= AR [sin (25°)]
= 6.104 × 0.4226
= 2.58 cm
For violet colour, L_V = VM
= AV [sin (i – r_V)]
= AV × sin (25° 32′)
= 6.065 × 0.431
= 2.61 cm
∴ L_VR = L_V – L_R = 2.61 – 2.58 = 0.03 cm
= 0.3 mm

Question 41.
For a glass (n = 1.5) prism having refracting angle 60°, determine the range of angle of incidence for which emergent ray is possible from the opposite surface and the corresponding angles of emergence. Also calculate the angle of incidence for which i = e. How much is the corresponding angle of minimum deviation?
Answer:
For an equilateral prism of glass, the minimum angle of incidence for which the emergent ray just emerges is i_min = 27° 55′. Corresponding angle of emergence is, e_max = 90°.
From the principle of reversibility of light, i_max = 90° and e_min = 27°55′
Also, for equilateral glass prism at minimum deviation,

Also, from prism formula,
i + e = A + δ
At minimum deviation,
∴ i + i = 60 + 37°10′ = 97°10′
∴ i = 48°35′

Question 42.
For a dense flint glass prism of refracting angle 10°, obtain angular deviation for extreme colours and dispersive power of dense flint glass. (n_red = 1.712, n_violet = 1.792)
Answer:
Given: A = 10°, n_R = 1.712, n_V = 1.792
To find:
i. Angular deviation for extreme colours (δ_V and δ_R)
ii. Dispersive power of flint glass (ω)
Formulae:
i. δ = A(n – 1)
ii. ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)

Question 43.
The refractive indices of the material of the prism for red and yellow colour are 1.620 and 1.635 respectively. Calculate the angular dispersion and dispersive power, if refracting angle is 8°.
Solution:
Given: n_R = 1.620, n_Y = 1.635, A = 8°
To find:
i. Angular dispersion (δ_V – δ_R)
ii. Dispersive power (ω)
Formulae:
i. δ_v – δ_r = A(n_V – n_R)
ii. ω = \(\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)
Calculation: Since, n_Y = \(\frac {n_V+n_R}{2}\)
∴ n_V = 2n_Y – n_R
n_V = 2 × 1.635 – 1.620 = 3.27 – 1.620
∴ n_V = 1.65
From formula (i),
δ_V – δ_R = 8(1.65 – 1.620)
= 8 × 0.03 = 0.24°
∴ δ_V – δ_R = 0.24°
From formula (ii),
ω = \(\frac{1.65-1.620}{1.635-1}=\frac{0.03}{0.635}\) = 0.0472

Question 44.
What could be the possible reasons for the upward bending of the light ray during hot days?
Answer:
Possible reasons for the upward bending at the road could be:
i. Angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large which causes reflection.
ii. Air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards and once it has bent upwards, it continues to do so.

Question 45.
State some properties of rainbow.
Answer:

1. It is seen during rains and on the opposite side of the Sun.
2. It is seen only during mornings and evenings and not throughout the day.
3. In the commonly seen rainbow red arch is outside and violet is inside.
4. In the rarely occurring concentric secondary rainbow, violet arch is outside and red is inside.
5. It is in the form of arc of a circle.
6. Complete circle can be seen from a higher altitude, i.e. from an aeroplane.
7. Total internal reflection is not possible in this case.

Question 46.
Why is total internal reflection not possible during formation of a rainbow ?
Answer:
i. For total internal reflection, the angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

ii. The relative refractive index of air with the water drop is just less than 1 and hence the critical angle is almost equal to 90°.

iii. Angle of incidence i in air, at the water drop, can’t be greater than 90°.

iv. As a result, angle of refraction r in water will always be less than the critical angle.

v. The figures shown indicates that this angle r itself acts as an angle of incidence at any point for one or more internal reflections. But this does not indicate the total internal reflection.

Question 47.
Rainbow is seen only for a definite angle range with respect to the ground. Justify.
Answer:
i. For clear visibility of rainbow, a beam must have enough intensity.

ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity.

iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

Question 48.
How is the range of angles for which rainbows can be observed calculated?
Answer:
i. Angle of deviation for the final emergent ray, can be shown to be equal to δ = π + 2i – 4r for primary rainbow and δ = 2π + 2i – 6r for the secondary rainbow.
ii. Using these relations along with Snell’s law, sin i = n sin r, derivatives of angle of deviation (δ) is obtained.
iii. Second derivative of δ comes out to be negative, which shows that it is the minima condition.
iv. Equating first derivative to zero corresponding values of i and r are obtained. Thus, from the figures shown, the corresponding angles θ_R and θ_V at the horizontal are obtained. These angles give the visible angular position for the rainbow.

Question 49.
When can one see complete circle of a rainbow? Explain in detail.
Answer:
i. Figure given below illustrates formation of primary and secondary rainbows with their common centre O. It is the point where the line joining the sun and the observer meets the Earth when extended.

ii. P is location of the observer. Different colours of rainbows are seen on arches of cones of respective angles.

iii. Smallest half angle refers to the cone of violet colour of primary rainbow, which is 41°.

iv. As the Sun rises, the relative position of common centre of the rainbows with respect to observer shifts down. Hence as the Sun comes up, smaller and smaller part of the rainbows will be seen. If the Sun is above 41°, violet arch of primary rainbow cannot be seen.

v. Beyond 53°, no rainbows will be visible. That is why rainbows are visible only during mornings and evenings and in the shape of a bow.

vi. However, if observer moves up (may be in an aeroplane), the line PO itself moves up making lower part of the arches visible. After a certain minimum elevation, entire circle for all the cones can be visible.

Question 50.
Define following terms:
i. Longitudinal chromatic aberration
ii. Circle of least confusion
iii. Transverse chromatic aberration
Answer:
i. Longitudinal chromatic aberration:
Due to different refractive indices and angle of deviations, violet and red colours of a white light converge at different focal points, f_V and f_R. The distance between f_V and f_R is measured as the longitudinal chromatic aberration.

ii. Circle of least confusion:
In presence of aberration the image is not a single point but always a circle. At particular location on the screen, this circle has minimum diameter. This is called circle of least confusion.

iii. Transverse chromatic aberration:
Radius of the circle of least confusion is called the transverse chromatic aberration.

Question 51.
After Cataract operation, a person is recommended with concavo-convex spectacles of curvatures 10 cm and 50 cm. Crown glass of refractive indices 1.51 for red and 1.53 for violet colours is used for this. Calculate the lateral chromatic aberration occurring due to these glasses.
Answer:
For a concavo-convex lens, with convex shape facing the object, both the radii of curvature are positive as shown in the figure.

= (1.53 – 1) × 0.08 = 0.0424
∴ f_v = 23.58 cm
∴ Longitudinal (lateral) chromatic aberration
= f_V – f_R = 24.51 – 23.58 = 0.93 cm

Question 52.
Why do we need optical instruments for?
Answer:
i. Due to the limitation for focusing the eye lens it is not possible to take an object closer than a certain distance. This distance is called least distance of distance vision D. For a normal, unaided human eye D = 25cm.
ii. If an object is brought closer than this, we cannot see it clearly.
iii. If an object is too small the corresponding visual angle from 25 cm is not enough to see it and if we bring it closer than that, its image on the retina is blurred.
iv. Also, the visual angle made by cosmic objects far away from us such as stars is too small to make out minor details and we cannot bring those closer.
v. In such cases we need optical instruments like microscope and telescopes to observe these things clearly.

Question 53.
A convex lens has focal length of 2.0 cm. Find its magnifying power if image is formed at DDV.
Answer:
Given: f = 2 cm, v = D = 25 cm
To find: Magnifying power (M.P.)
Formula: M.P = 1 + \(\frac {D}{f}\)
Calculation:
From formula,
M.P = 1 + \(\frac {25}{2}\)
M.P. = 1 + 12.5 = 13.5

Question 54.
A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size. How big will the letters appear? Can you read the letters if held 5 cm away from the lens? If yes, of what size would the letters appear? If no, why not?
Answer:
Given that, f = +10 cm, u = -8 cm,
From thin lens formula,
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
∴ \(\frac{1}{10}=\frac{1}{\mathrm{v}}-\frac{1}{-8}\)
∴ v = 40 cm
Magnification of a lens is,
m = \(\frac {v}{u}\) = \(\frac {Object size h-i}{Object size h-0}\)
∴ \(\frac {40}{8}\) = \(\frac {h_1}{0.5}\)
∴ h₁ = 2.5 cm
This implies the height of the image is 5 times that of the object.
Magnifying power,
M = \(\frac {D}{u}\) = \(\frac {25}{8}\) = 3.125
∴ Image will appear to be 3.125 times bigger,
i.e., 3.125 × 0.5 = 1.5625 cm
For u = -5 cm, v will be -10 cm
For an average human being to see clearly, the image must be at or beyond 25 cm. Thus it will not possible to read the letters if held 5 cm away from the lens.

Question 55.
A compound microscope has a magnification of 15. If the object subtends an angle of 0.5° to eye, what will be the angle subtended by the image at the eye?
Answer:
Given: M.P = 15, α = 0.5°
To Find: Angle(β)
Formula: M.P = \(\frac {β}{α}\)
Calculation:
From formula,
β = M.P × α = 15 × 0.5 = 7.5°

Question 56.
A compound microscope has a magnifying power of 40. Assume that the final image is formed at DDV(25 cm). If the focal length of eyepiece 10 cm, calculate the magnification produced by objective.
Answer:
M.P = 40, D = 25 cm, f_e = 10 cm
To Find: Magnification (m₀)
Formula: M.P = m₀ × M_e
Calculation:
From the formula,

Question 57.
The pocket microscope used by a student consists of eye lens of focal length 6.25 cm and objective of focal length 2 cm. At microscope length 15 cm, the final image appears biggest. Estimate distance of the object from the objective and magnifying power of the microscope.
Answer:
Given: f_e = 6.25 cm, f₀ = 2 cm, L = 15 cm
As image appears biggest, V_e = -25 cm.
To find:
i. Distance of object from objective (u₀)
ii. Magnifying power (M)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. L = |v₀| + |u_e|
iii. M = \(\left(\frac{v_{o}}{u_{o}}\right)\left(\frac{D}{u_{e}}\right)\)
Calculation: For eyelens, using formula (i),

∴ u_e = 5 cm
From formula (ii),
|v₀| = L – |u_e|
= 15 – 5 = 10 cm
Using formula (i) for objective,

Question 58.
Focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, length of the telescope is 1.05 m. Calculate focal length of the eyepiece and magnifying power under normal adjustment.
Answer:
Given: f₀ = 1 m, L = 1.05 m
To find:
i. Focal length of eyepiece (f_e)
Magnifying power under normal adjustment (M)
Formula:
i. L = f₀ + f_e
ii. M = \(\frac {f_0}{f_e}\)
Calculation. From formula (i),
f_e = L – f₀ = 1.05 – 1 = 0.05 m
From formula (ii),
M = \(\frac {1}{0.05}\) = 20

Question 59.
Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
Answer:
Given: M.P = 12, v = D, f₀ = 90 cm,
To find: Focal length of eye piece (f_e)
Formula: M.P = \(\frac {f_0}{f_e}\) (1 + \(\frac {f_e}{D}\))
Calculation:
From formula.
12 = \(\frac {90}{f_e}\) (1 + \(\frac {f_e}{25}\))
∴ f_e = 10.71 cm

Question 60.
Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length of the telescope.
Answer:
Given: f₀ = 1.3 m, f_e = 0.05 m
To find:
i. Magnifying power of telescope (M.P.)
ii. Length of telescope (L)
Formulae:
i. M.P = \(\frac {f_0}{f_e}\)
ii. L = f₀ + f_e
Calculation: From formula (i),
M.P = \(\frac {1.3}{0.05}\) = 26
From formula (ii),
L = 1.3 + 0.05 = 1.35 m

Question 61.
What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?
Answer: When a ray of light is incident on a plane mirror at an angle θ, the reflected ray gets deviated by an angle of (π – 2θ).

Question 62.
Does nature of the image depend upon size of the mirror?
Answer:
No, nature of the image is independent of size of the mirror.

Question 63.
If a convex mirror is held in air and then dipped in oil, then what will be the change in its focal length?
Answer:
Focal length of spherical mirrors are independent of the medium.

Question 64.
When ray of light falls normally on a mirror, its angle of incidence is 90°. True or false? Justify your answer.
Answer:
False, when light falls normally on a mirror, its angle of incidence is zero degree.

Question 65.
In one of the performances, a magician keeps a gold ring beneath a thick glass slab (µ = \(\frac {3}{2}\)) Then he keeps a flask filled with water (µ = \(\frac {4}{3}\)), over the slab. Now when spectators one by one observe from the open end of the flask, the ring disappears at a certain angle of viewing.
i. What could be the reason behind the disappearance?
ii. At what angle of viewing the ring vanishes?
Answer:
i. The ring disappears due to total internal reflection of the light at water-air interface.

ii. _aµ_w = \(\frac {4}{3}\)
sin i_c = \(\frac {1}{µ}\)
∴ i_c = sin^-1 (\(\frac {1}{_aµ_w}\)) = sin^-1 (\(\frac {3}{4}\)) = 48.6°
Hence, for angle of viewing for which angle of incidence of ring from water to air is greater that 48.6°, the ring will vanish.

Question 66.
Why dispersion of light is not observed in glass slab but it is observed in prism?
Answer:
When a light passes from one medium to another, at one interface, it changes its speed. The glass slab and prism both have two glass-air interfaces. Hence, the light undergoes refraction twice in both the cases. When the two interfaces are parallel to each other, although the colours are separated at first interface, they all travel the same path after refracting from second interface. However, in prism, the two interfaces are not parallel. Therefore, the colours separated at first interface do not travel the same path after second refraction but emerge out at different wavelengths producing spectrum.

Question 67.
A prism manufacturer is planning to build a dispersive prism out of following materials with the refracting angles as given.
i. Glass (µ = 1.5), A = 60°
ii. Plastic (µ = 1.4), A = 90°
iii. Fluorite (µ = 1.45), A = 64°
If he desires to give the prism following relations of i and δ, then which of the above combinations can be used to construct the prism?

Answer:
From the given values of i and δ, δm = 37°
From prism formula, µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)

i. For Glass (µ = 1.5), A = 60°;
µ = \(\frac{\sin \left(\frac{60^{\circ}+37^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= 1.5
Hence, this combination can be used for fabricating the desired prism.

ii. For Plastic (µ = 1.4), A = 90°;
µ = \(\frac{\sin \left(\frac{90^{\circ}+37^{\circ}}{2}\right)}{\sin \left(45^{\circ}\right)}\)
= 1.26
As P_piastic = 1-4, this combination cannot be used.

iii. For Fluorite (µ = 1.45), A = 64°
µ = \(\frac{\sin \left(\frac{64^{\circ}+37^{\circ}}{2}\right)}{\sin \left(32^{\circ}\right)}\)
= 1.45
Hence, this combination can also be used for fabrication of prism.

Question 68.
Find the refractive index of material of following prism if the ray of light incident at angle 45° suffers minimum deviation through the prism.

Answer:

A = 60°
Also, ray of light suffers minimum deviation.
∴ 2i = A + δ_m
∴ δ_m = 2i – A = 90° – 60° = 30°
From prism formula,
µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
= \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= √2
Hence, refractive index of material of prism is √2.

Multiple Choice Questions

Question 1.
Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
(A) 4 × 10^-11 s
(B) 2 × 10^-11 ns
(C) 16 × 10^-11 s
(D) 8 × 10^-10 s
Answer:
(A) 4 × 10^-11 s

Question 2.
If mirrors are inclined to each other at an angle of 90°, the total number of images seen for a symmetric position of an object will be
(A) 3
(B) 4
(C) 5
(D) 3 or 4
Answer:
(A) 3

Question 3.
In case of a convex mirror, the image formed is
(A) always on opposite side, virtual, erect.
(B) always on the same side, virtual, erect.
(C) always on opposite side, real, inverted.
(D) dependent on object distance.
Answer:
(A) always on opposite side, virtual, erect.

Question 4.
A glass slab is placed in the path of a beam of convergent light. The point of convergence of light
(A) moves towards the glass slab.
(B) moves away from the glass slab.
(C) remains at the same point.
(D) undergoes a lateral shift.
Answer:
(A) moves towards the glass slab.

Question 5.
For a person seeing an object placed in optically rarer medium,
(A) apparent depth of the object is more than real depth
(B) apparent depth is smaller than the real depth.
(C) apparent depthe might be smaller or greater depending on the position of the person.
(D) nothing can be concluded about the depth of object from given data.
Answer:
(A) apparent depth of the object is more than real depth

Question 6.
Light travels from a medium of refractive index µ₁ to another of refractive index µ₂ (µ₁ > µ₂). For total internal reflection of light, which is NOT true?
(A) Light must travel from medium of refractive index µ₁ to µ₂.
(B) Angle of incidence must be greater than the critical angle.
(C) There is no refraction of light.
(D) Light must travel from the medium of refractive index µ₂ to µ₁.
Answer:
(D) Light must travel from the medium of refractive index µ₂ to µ₁.

Question 7.
Optical fibre is based on which of the following phenomenon?
(A) Reflection.
(B) Refraction.
(C) Total internal reflection.
(D) Dispersion.
Answer:
(C) Total internal reflection.

Question 8.
Commonly used glass have refractive index of 1.5. What is the critical angle for such glass?
(A) 49°
(B) 42°
(C) 45°
(D) 40°
Answer:
(B) 42°

Question 9.
If the refractive index of water is 4/3 and that of glass slab is 5/3. Then the critical angle of incidence for which a light ray tending to go from glass to water is totally reflected, is
(A) sin^-1 (\(\frac {3}{4}\))
(B) sin^-1 (\(\frac {3}{5}\))
(C) sin^-1 (\(\frac {2}{3}\))
(D) sin^-1 (\(\frac {4}{5}\))
Answer:
(D) sin^-1 (\(\frac {4}{5}\))

Question 10.
While deriving prism formula, which of the following condition is NOT satisfied?
(A) I = e
(B) r₁ = r₂
(C) r = \(\frac {A}{2}\)
(D) δ_m = i + e + r
Answer:
(D) δ_m = i + e + r

Question 11.
If the critical angle for the material of a prism is C and the angle of the prism is A, then there will be no emergent ray when
(A) A < 2 C
(B) A = 2 C
(C) A > 2 C
(D) A < \(\frac {C}{2}\)
Answer:
(C) A > 2 C

Question 12.
Chromatic aberrations is caused due to
(A) spherical shape of lens
(B) spherical shape of mirrors
(C) angle of deviation for violet light being more than that for red light.
(D) refractive index for violet light being less than that for red light
Answer:
(C) angle of deviation for violet light being more than that for red light.

Question 13.
In normal adjustment, magnifying powser of a astronomical telescope is given by
(A) \(\frac {D}{f_0}\) \(\frac {L}{f_e}\)
(B) \(\frac {L}{D}\) \(\frac {f_e}{f_0}\)
(C) \(\frac {f_0}{f_e}\)
(D) \(\frac {f_e}{f_0}\)
Answer:
(C) \(\frac {f_0}{f_e}\)

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 12 Photosynthesis Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 12 Photosynthesis

Question 1.
What is the percentage of C0₂ in atmosphere?
Answer:
Atmosphere contains only about 0.03 percent C0₂ by volume.

Question 2.
Define photosynthesis and give its reaction with the help of solar energy trapped by pigments like chlorophyll.
Answer:
Photosynthesis is defined as synthesis of carbohydrates (glucose) from inorganic materials like CO₂ and H₂0 with the help of solar energy trapped by pigments like chlorophyll.

Question 3.
Explain diagrammatically the ultrastructure of chloroplast.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
2. System of chlorophyll bearing a double-membrane sac is present inside the stroma. These are stacked one above the other to form grana.
3. Individual sacs in each granum is are known as thylakoid.
4. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
5. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Question 4.
Distinguish between grana and stroma.
Answer:

Grana	Stroma
1. These are formed of stacks of thylakoids.	It is the matrix of chloroplast.
2. Light reaction occurs in grana.	Dark reaction occurs in stroma.

Question 5.
Describe a structure of a chlorophyll molecule.
Answer:
1. Chemically chlorophyll molecule consists of two parts head of tetrapyrrole the porphyrin ring and a long hydrocarbon tail called phytol attached to the porphyrin group.
2. Both chlorophyll-a and chlorophyll-b are similar in their molecular structure, except that the methyl group (-CH₃) in chlorophyll-a is replaced with an aldehyde group (-CH_O) in chlorophyll-b.

Question 6.
What are Carotenoids?
Answer:

1. Carotenoids are lipid compound present universally in almost all the higher plants and several microorganisms.
2. They are usually red, orange, yellow, brown, and are associated with chlorophyll. They are of two types – the carotenes and xanthophylls.
3. The carotenes (C₄₀H₅₆) are orange red and xanthophylls contain oxygen.
4. The light energy absorbed by the carotenoids is transferred to chlorophyll-a to be utilized in photosynthesis.
5. All photosynthetic plants have these pigments that absorb light between the red and blue region of the
   spectrum.
6. Carotenoids found mainly in higher plants, absorb primarily in the violet to blue regions of the spectrum.
7. They not only absorb light energy and transfer it to chlorophyll but also protect the chlorophyll molecule from photo-oxidation.

Question 7.
Write a short note on nature of light.
Answer:

1. Light is a form of energy.
2. It travels as stream of tiny particles called photons.
3. A photon contains a quantum of light.
4. Light has different wavelengths having different colors.
5. One can see electromagnetic radiation with wavelengths ranging from 390nm to 730nm. This part of the spectrum is called the Visible light.
6. It lies between wavelengths of ultraviolet and infra-red.

Question 8.
Draw neat and labelled diagram of absorption and action spectrum.
Answer:
Absorption and action spectra

Question 9.
Write the chemical reaction given by Van Niel in case of bacteria that use H₂S and C0₂ to synthesize carbohydrates. Write his postulate.
Answer:
Chemical reaction given by Van Niel:
6C0₂ + 12H₂S → C₆H₁₂0₆ + 6H₂0 + 12S ↓
Van Niel’s postulate:
Green plants use water in place of H₂S and 0₂ is evolved in place of sulphur.

Question 10.
What was used by Ruben to confirm that the source of oxygen evolved in photosynthesis is water?
Answer:
Ruben used heavy isotope of oxygen (¹⁸0₂) to confirm that the source of oxygen evolved in photosynthesis is water.

Question 11.
Explain Hill reaction.
Answer:
In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.

Question 12.
Give reason. Why photosynthesis is a redox reaction?
Answer:
Photosynthesis is considered as a redox reaction as it involves both reduction and oxidation reactions.
Water is oxidized by the removal of H⁺ while C0₂ is reduced by the addition of H⁺.
The redox reactions of photosynthesis are necessary for the conversion of light energy into chemical energy.

Question 13.
Which are the two reactions involved in the process of photosynthesis.
Answer:

1. The process of photosynthesis is an oxidation and reduction process in which water is oxidized (to release 0₂) and C0₂ is reduced to form sugar. It consists of two successive series of reactions.
2. The first reaction requires light and is called Light reaction.
3. Second reaction does not require light and is called Dark or Blackman reaction.
4. Of the two reactions, the former is a photochemical reaction, while the latter is a biochemical reaction.

Question 14.
What is light reaction?
Answer:
1. The light reaction is a reaction in which solar energy is trapped by chlorophyll and stored in the form of chemical energy as ATP and in the form of reducing power as NADPH₂.
2. Oxygen is evolved in the light reaction by splitting of water.

Question 15.
Define: Quantum energy
Answer:
A certain minimum quantity of energy a photon must have to boost an electron is called quantum energy.

Question 16.
What is excited state?
Answer:
A molecule that has absorbed a photon is in energy rich excited state. An excited state of an atom means that the valence electron has moved from its ground state orbital to high energy orbital.

Question 17.
What is ground state?
Answer:
When the light source is turned off, the high energy electrons return rapidly to their normal low energy orbitals as the excited molecule reverts to its original stable condition, called the ground state.

Question 18.
Explain the structure of reaction centre.
Answer:
Solar energy is trapped by chlorophyll and stored in the form of chemical energy as ATP and in the form of reducing power as NADPH₂. Oxygen is evolved in the light reaction by splitting of water. The components of light reaction are as follows:

Reaction centre:

1. The light absorbing pigments present in thylakoid membranes are arranged in clusters of chlorophyll and accessory pigments.
2. P-680 and P-700 are the special type of chlorophyll molecules which form the reaction centres or photocentres.
3. Solar energy is harvested by accessory pigments and other chlorophyll molecule and is passed on to the reaction centre.
4. These (accessory pigments) are known as light harvesting or antenna molecule. Their function is to absorb light energy and transmit at a very high rate to the reaction centre where the photochemical reactions occur.

Question 19.
What are the two types of photosystems?
Answer:
1. Photosystems:
a. Two kinds of photosystems are present in thylakoid membranes of chloroplasts.
b. Each has its own set of light harvesting chlorophyll and carotenoid molecules.
c. Chlorophyll and accessory pigments help to capture light energy over large area and pass it on to the photocenters.
d. Thus, a photon absorbed anywhere in the harvesting zone of P-680 center can pass its energy to the P680 molecule.
e. The cluster of pigments which transfer their energy to P-680 absorb at or below 680nm.
f. Together with P-680 they form Photosystem – II.
g. Likewise, P-700 forms Photosystem – I along with pigment molecule which absorbs light at or below 700 nm.

2. Photosystem II:
a. Photolysis of water and release of oxygen takes place in this system.
b. In this process, when PS-II absorbs light, electrons are released and chlorophyll molecule is oxidized.
c. Electrons emitted by P680 (PS-II) are ultimately trapped by P700 (PS -1).
d. Oxygen is the byproduct by the photosynthesizing plants.
e. Protons accumulate inside the thylakoid resulting in a Proton gradient.
f. When the protons diffuse across the thylakoid membrane into stroma against the H+ gradient, energy, is released.
g. This is used to produce ATP.

3. Photosystem 1:
a. Upon absorption of light quanta by PS-I (P700) reaction center emits energy rich electrons.
b. These flow down a chain of electron carriers to NADP along with the proton generated by splitting of water.
c. This result in the formation of NADPH.
d. Hydrogen attached to NADPH is used for reduction of C0₂ in dark reaction also called as reducing power of the cell.

Question 20.
The oxidized P-680 regains its electrons by the photolysis of water. Mention the reactions related to the same.
Answer:
The oxidized P-680 regains its electrons by the photolysis of water as follows:

1. 4H₂O → 4H⁺ + 40H^–
2. 4OH^– → 4(OH) + 4e^–
3. 4OH^– → 2H₂0 + 0₂
4. 4H₂O → 2H₂O + 0₂ ↑ + 4H⁺ 4e^– Overall reaction

Question 21.
Distinguish between Photosystem I and Photosystem II
Answer:

1. The movement of ions across a selectively permeable membrane, down their electrochemical gradient is called chemiosmosis.
2. The movement of hydrogen ions across a membrane during cellular respiration in mitochondria and during photosynthesis in chloroplasts, leads to the generation of ATP.
3. These membranes are thylakoids and the protons accumulate in the lumen during photosynthesis.
4. An electrochemical concentration gradient forms when hydrogen ions (protons) diffuse from an area of higher proton concentration to an area of lower proton concentration.
5. This electrochemical concentration gradient of protons across a membrane can be utilized to make
   ATP.
6. ATP synthase is the enzyme that makes ATP by chemiosmosis.
7. It allows protons to pass through the membrane using the kinetic energy to phosphorylate ADP making ATP.
8. Splitting of water molecule on the inner side of the membrane results in the accumulation of hydrogen ions within the lumen of thylakoids.
9. The enzyme, NADP reductase, is located in the stroma side of the membrane.
10. For reduction of NADP to NADPH₂, protons are required along with electrons that come from ferredoxin.
11. Thus, within the chloroplast, the protons in the stroma decrease in number, while in the lumen, the number of protons increases.
12. This creates a proton gradient across the thylakoid membrane.
13. Energy,generated by the subsequent spontaneous movement of protons is used for the synthesis of ATP.

Question 22.
Draw neat and labelled diagram representing ATP synthesis through chemiosmosis.
Answer:

Question 23.
What is photophosphorylation?
Answer:
Formation of ATP in the chloroplasts in presence of light is called photophosphorylation, light

Question 24.
Mention the light reactions occurring in in granum.
Answer:

1. 24H₂O → 24OH_– + 24H₊
2. 24OH_– → 240H + 24e_–
3. 24e^– + 24H₊ + 12NADP₊ → 12NADPH₂
4. 18ADP + 18Pi → 18 ATP
5. 240H → 12H₂O + 6O₂ ↑

Question 25.
Write the dark reaction occurring in stroma.
Answer:
60₂+ 18ATP + 12NADPH₂ → C₆H₁₂O₂ + 6H₂O + 18ADP +18Pi+ 12NADP⁺

Question 26.
Name the following:
1. Name the first compound formed during C₃ pathway.
2. Carbon fixation occurs in?
Answer:
1. 3-phosphoglyceric acid.
2. Stroma

Question 27.
Where does this reaction take place?
6C0₂ + 18ATP + 12NADPH₂ → C₆H₁₂O₆ + 6H₂0 + 18ADP + 18Pi + 12NADP⁺
Answer:
In stroma of a chloroplast.

Question 28.
Who discovered the Carbon fixation pathway? How was the pathway studied?
Answer:

1. The path of carbon fixation in dark reaction through intermediate compounds leading to the formation of sugar and starch was discovered by Calvin, Benson and their co-workers discovered the carbon fixation pathway.
2. Path of carbon was studied with the help of radioactive tracer technique using Chlorella, a unicellular green alga and radioactive ¹⁴C0₂.
3. With the help of radioactive carbon, it becomes possible to trace the intermediate steps of fixation of ¹⁴C0₂.

Question 29.
When does photorespiration occur?
Answer:
1. Photorespiration occurs under the conditions like high temperature, bright light, high oxygen and low C0₂ concentration. It is a wasteful process linked with C₃-Cycle, where instead of fixation of C0₂, it is given out.
2. Photorespiration is a respiratory process in many higher plants by which they take up oxygen in the light and give outiearbon dioxide.

Question 30.
Diagrammatically explain the process of photorespiration.
Answer:
Mechanism:

1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized in formation of 3-PGA, which get utilized in C₃ pathway.

Question 31.
Explain Hatch – Slack pathway.
Answer:
1. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
2. It occurs in tropical and sub-tropical grasses and some dicotyledons.
3. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP. Thus, C₄ pathway needs 12 additional ATP.
11. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
12. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

Question 32.
Draw a neat labelled diagram of Kranz anatomy of C₄ plant.
Answer:

Question 33.
Name the following:
1. Name the first product of C0₂ fixation in Hatch – Slack Pathway.
2. C₄ plant leaf shows which type of anatomy?
3. Example of a C₄ Plant
Answer:
1. Oxaloacetic acid.
2. Kranz Anatomy
3. Maize, Sugarcane, Sorghum, etc.

Question 34.
Short note on Kranz Anatomy of a C₄ plant.
Answer:

1. Anatomy of leaves of C₄ plants is different from leaves of C₃ plants.
2. C₄ plants show Kranz anatomy.
3. In the leaves of such plants, there is a bundle sheath around the vascular bundles.
4. The chloroplasts in the bundle – sheath cells are large and without or less developed grana, where as in the mesophyll cells the chloroplasts are small but with well-developed grana.

Question 35.
Short note on: Crassulacean Acid Metabolism.
Answer:
Crassulacean Acid Metabolism (CAM).

1. It is one more alternative pathway of carbon fixation found in desert plants.
2. It was first reported in the family Crassulaceae, therefore called as CAM (Crassulacean Acid Metabolism).
3. In CAM plants, stomata are scotoactive i.e. active during night, hence initial C0₂ fixation occurs in night.
4. Thus, C₄ pathway fix C0₂ at night and reduce C0₂ in day time via the C₃ pathway by using NADPH formed during the day.
5. PEP caboxylase and Rubisco are present in the mesophyll cell (no Kranz anatomy).
6. Formation of malic acid during dark is called acidification (phase I).
7. Malate is stored in vacuoles during the night.
8. Malate releases C0₂ during the day for C₃ pathway within the same cell is called deacidification (phase II).
9. Examples of CAM plants: Kalanchoe, Opuntia, Aloe etc.
10. The Chemical reactions of the carbon dioxide fixation and its assimilation are similar to that of C₄ plants.

Question 36.
1. Draw a neat labelled diagram of CAM pathway.
2. In CAM plants, why does acid concentration increase during night?
Answer:

2. In CAM plants, malic acid accumulates during night, which is formed from Oxaloacetic acid in presence of the enzyme malate dehydrogenase.

Question 37.
What are the external factors which affects photosynthesis?
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.

Question 38.
What are the internal factors which affects photosynthesis?
Answer:
Internal factors which affects photosynthesis are as follows:
1. Chlorophyll:
a. Though presence of chlorophyll is essential for photosynthesis but rate of photosynthesis is proportional to the quantity of chlorophyll present.
b. It is because of the fact that chlorophyll merely acts as a biocatalyst and hence a small quantity is quite enough to maintain the large bulk of the reacting substances.
2. Sugar:
The final product in the photosynthesis reaction is sugar and its accumulation in the cells slow down the process of photosynthesis.
3. Internal structures:
The thickness of cuticle and epidermis of the leaf, the size and distribution of intercellular spaces and the distribution of the stomata and the development of chlorenchyma and other tissues also affects the rate of photosynthesis.

Question 39.
State and explain the Blackman’s law of limiting factor.
Answer:

1. The Blackman’s law of limiting factors states that when a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is controlled by the pace of the “slowest factor”.
2. The slowest factor is that factor which is present in the lowest or minimum concentration in relation to others.
3. The law of limiting factor can be explained by taking two external factors such as carbon dioxide and light.
4. For example, a plant photosynthesizing at a fixed light intensity sufficient to utilize 10mg of C₂ per hour only.
5. Photosynthetic rate goes on increasing when concentration of CO₂ increases.
6. Further increase in CO₂ concentration will not increase the rate of photosynthesis. In this case light becomes the limiting factor. Therefore, under such circumstances rate of photosynthesis can be increased only by increasing the light intensity.
7. This proves that the rate of photosynthesis responds to one factor alone at a time and there would be a sharp break in the curve and a plateau formed exactly at the point where another factor becomes limiting.
8. If any one of the other factors which is kept constant (e.g. Light) is increased, the photosynthetic rate increases again reaching the optimum where again another factor becomes limiting.

Question 40.
Give the significance of Photosynthesis.
Answer:

1. Photosynthesis is anabolic process which uses inorganic substances and produces food for all life directly or indirectly.
2. This process transforms solar energy into chemical energy.
3. The released by product 0₂ is necessary not only for aerobic respiration in living organisms but also used in forming protective ozone layer around earth.
4. It also helps us in providing fossil fuels, coals, petroleum and natural gas.

Question 41.
Apply Your Knowledge:

Question 1.
1. Identify ‘x’, ‘y’ and ‘z’ in the given diagrammatic representation of cyclic photophosphorylation.
2. During which steps ATP molecules are formed?
Answer:

1. ‘x’ is ferredoxin, ‘y’ is cytochrome-b6 and ‘z’ is plastocyanin.
2. During conversion of ferredoxin to cytochrome-b6 and from cytochrome-b6 to cytochrome-f ATP molecules are formed.

Question 42.
Quick Review

Cyclic and Non-cyclic photophosphorylation:

	Cyclic	Non-cyclic
Used	Photons 1 ADP 1 Phosphate group	Photons 1 ADP1 Phosphate group 1 H20 1 NADP+
Produced	2 ATP	1 ATP 2 NADPH + H+ \(\frac { 1 }{ 2 }\)O2
Accomplished	Captured energy in the form of ATP	Captured energy in the form of ATP and NADPH2 Transfers hydrogen (as NADPH) to the dark reactions

Calvin’s Cycle:

Used (Reactants)	Produced (Products)
6CO2	12 PGAL (2 become 1 glucose)
6 RUBP	18 Phosphates (return to light reactions)
18 ATP (from light reactions)	18 ADP (returns to light reactions)
12 NADPH (from light reactions)	12 NADP+ (return to light reactions)

Abbrevations

ATP	Adenosine triphosphate	RuBisCO	Ribulose bisphosphate carboxylase
ADP	Adenosine diphosphate	PGA	Phosphoglyceric acid
Co-Q	Co-enzyme quinone	PGAL	Phosphoglyceraldehyde
FRS	Ferredoxin Reducing Substance	DHAP	Dihydroxyacetone phosphate
PC	Plastocyanin	NADP	Nicotinamide adenine dinucleotide phosphate
PQ	Plastoquinone	NADPH2	Nicotinamide adenine dinucleotide hydrogen phosphate
RUBP	Ribulose-1, 5-bisphosphate	PEPA	Phosphoenol pyruvic acid
RUMP	Ribulose monophosphate	OAA	Oxaloacetic acid

Question 43.
Exercise:

Question 1.
Why photosynthesis is known as redox reaction?
Answer:

1. Photosynthesis is considered as a redox reaction as it involves both reduction and oxidation reactions.
2. Water is oxidized by the removal of H₊ while C0₂ is reduced by the addition of H.
3. The redox reactions of photosynthesis are necessary for the conversion of light energy into chemical energy.

Question 2.
Sketch and label ‘Ultrastructure of Chloroplast’.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
2. System of chlorophyll bearing a double-membrane sac is present inside the stroma.
3. These are stacked one above the other to form grana.
4. Individual sacs in each granum is are known as thylakoid.
5. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
6. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Question 3.
Name the various photosynthetic pigments.
Answer:

1. Chemically chlorophyll molecule consists of two parts head of tetrapyrrole the porphyrin ring and a long hydrocarbon tail called phytol attached to the porphyrin group.
2. Both chlorophyll-a and chlorophyll-b are similar in their molecular structure, except that the methyl group (-CH) in chlorophyll-a is replaced with an aldehyde group (-CHO) in chlorophyll-b.
3. Carotenoids are lipid compound present universally in almost all the higher plants and several microorganisms.
4. They are usually red, orange, yellow, brown, and are associated with chlorophyll. They are of two types – the carotenes and xanthophylls.
5. The carotenes (C₄0H₅) are orange red and xanthophylls contain oxygen.
6. The light energy absorbed by the carotenoids is transferred to chlorophyll-a to be utilized in photosynthesis.
7. All photosynthetic plants have these pigments that absorb light between the red and blue region of the
   spectrum.
8. Carotenoids found mainly in higher plants, absorb primarily in the violet to blue regions of the spectrum.
9. They not only absorb light energy and transfer it to chlorophyll but also protect the chlorophyll molecule from photo-oxidation. Xanthophylls (C₄0H₅6O₂) are yellow pigments found in fruits and vegetables.

Question 4.
What is the main function of accessory pigment?
Answer:
1. Accessory pigments are light absorbing molecules which are found in photosynthetic organisms.
2. They transfer the absorbed light to chlorophyll-a and thus increasing the photosynthetic rate.
3. In absence of accessory pigments less amount of light will be absorbed and also there would be no protection provided to chlorophyll molecule from photo-oxidation.

Question 5.
Explain the nature of light.
Answer:

1. Light is a form of energy.
2. It travels as stream of tiny particles called photons.
3. A photon contains a quantum of light.
4. Light has different wavelengths having different colors.
5. One can see electromagnetic radiation with wavelengths ranging from 390nm to 730nm. This part of the spectrum is called the Visible light.
6. It lies between wavelengths of ultraviolet and infra-red.

Question 6.
Which instrument is used for studying the absorption spectrum of photosynthetic pigments?
Answer:
In absorption spectrum, absorption of different wavelengths of light pigments can be measured by spectrophotometer.

Question 7.
Explain Hill’s reaction.
Answer:
Robert Hill proved that the source of oxygen evolved during photosynthesis is water and not carbon dioxide. Hence, it is called Hill’s Reaction.

1. In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.
2. Ferric salts and haemoglobin were added in the medium as hydrogen and oxygen acceptors respectively.
3. When the suspension was illuminated, he observed that haemoglobin turned into oxyhaemoglobin (red colour).
4. This confirmed that water must have oxidized releasing 0₂, that reacted with haemoglobin. Reduction of ferric compound was also indicated by change in colour.
5. The H₂O molecule oxidized to evolve 0₂ as a by-product. Thus, Hill proved that the source of evolving 0₂ is H₂0 and not C0₂.
6. This process of splitting up of water molecules under the influence of light in the presence of chlorophyll is called Photolysis of water or Hill Reaction.

Question 8.
Describe photoexcitation of chlorophyll-a.
Answer:

1. Chlorophyll-a is an essential photosynthetic pigment as it converts light energy into chemical energy and acts as a reaction centre.
2. Initially, it lies at ground state or singlet state but when it absorbs or receives photons (solar energy), it gets activated and goes in excited state or excited second singlet state.
3. In the excited state, chlorophyll-a emits an electron. The emitted electron is energy rich, i.e. has extra amount of energy.
4. Due to the loss of electron (e_–), chlorophyll-a becomes positively charged. This is the ionized state.
5. Chlorophyll-a molecule cannot remain in the ionized state for more than 10‘9 seconds. Hence the photo-chemical reaction or electron transfer occurs very fast.
6. The energy rich electron is then transferred through various electron acceptors and donors (carriers).
7. During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy is converted into chemical energy in the form of ATP.

Question 9.
Write the wavelengths required for proper functioning of PS-I and PS-II respectively?
Answer:
1. Together with P-680 they form Photosystem – II.
2. Likewise, P-700 forms Photosystem – I along with pigment molecule which absorbs light at or below 700 nm.

Question 10.
Differentiate between PS-IandPS-II.
Answer:

1. The movement of ions across a selectively permeable membrane, down their electrochemical gradient is called chemiosmosis.
2. The movement of hydrogen ions across a membrane during cellular respiration in mitochondria and during photosynthesis in chloroplasts, leads to the generation of ATP.
3. These membranes are thylakoids and the protons accumulate in the lumen during photosynthesis.
4. An electrochemical concentration gradient forms when hydrogen ions (protons) diffuse from an area of higher proton concentration to an area of lower proton concentration.
5. This electrochemical concentration gradient of protons across a membrane can be utilized to make
   ATP.
6. ATP synthase is the enzyme that makes ATP by chemiosmosis. .
7. It allows protons to pass through the membrane using the kinetic energy to phosphorylate ADP making ATP.
8. Splitting of water molecule on the inner side of the membrane results in the accumulation of hydrogen ions within the lumen of thylakoids.
9. The enzyme, NADP reductase, is located in the stroma side of the membrane.
10. For reduction of NADP to NADPH₂, protons are required along with electrons that come from ferredoxin.
11. Thus, within the chloroplast, the protons in the stroma decrease in number, while in the lumen, the number of protons increases.
12. This creates a proton gradient across the thylakoid membrane.
13. Energy,generated by the subsequent spontaneous movement of protons is used for the synthesis of ATP.

Question 11.
Define photophosphorylation.
Answer:
Formation of ATP in the chloroplasts in presence of light is called photophosphorylation, light

Question 12.
Give graphic representation of cyclic photophosphorylation.
Answer:
Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

Question 13.
Give graphic representation of noncyclic photophosphorylation.
Answer:
Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO2 in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Question 14.
Write the link between light dependent and dark reactions.
Answer:
Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy-rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C02 is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Question 15.
Write the dark reaction occurring in stroma.
Answer:
60₂+ 18ATP + 12NADPH₂ → C₆H₁₂O₂ + 6H₂O + 18ADP +18Pi+ 12NADP⁺

Question 16.
Explain how light and dark reactions of photosynthesis are interdependent.
Answer:
Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy-rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Question 17.
Give schematic representation of Calvin Cycle.
Answer:
1. The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
2. Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer. It is also called synthesis phase or second phase of photosynthesis.

The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Question 18.
Describe Calvin cycle and its significance.
Answer:
Significance:
1. Carboxylation: RuBisCO is the most abundant enzyme in the world. It is responsible for fixing carbon in the form of C0₂ into sugar. As a result of Carboxylation, the first stable product of carbon fixation i.e. 3- PGA is synthesized.
2. Reduction/Glycolytic reversal: NADPH2 donates electrons to 1, 3-Bisphoshoglycerate to form 3- phosphoglyceraldehyde molecules. During this process ADP and NADP are generated which are used in light reaction.
3. Regeneration of RuBP: Some 3-phosphoglyceraldehyde molecules are involved in production of glucose while others are recycled to regenerate the 5-carbon compound RuBP which used to accept new carbon molecules. Thus, regeneration of RuBP is required for Calvin cycle to run continuously.

Question 19.
How was the carbon fixation pathway studied?
Answer:
1. Path of carbon was studied with the help of radioactive tracer technique using Chlorella, a unicellular green alga and radioactive ¹⁴C0₂.
2. With the help of radioactive carbon, it becomes possible to trace the intermediate steps of fixation of ¹⁴C0₂.

Question 20.
Describe photorespiration with the help of diagrammatic representation.
Answer:
Mechanism:

1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized information of 3-PGA, which get utilized in C₃ pathway.

Question 21.
Give significance of C₄ pathway.
Answer:

1. C₄ plants have special type of leaf anatomy called Kranz anatomy.
2. In C₄ plants, C0₂ fixation occurs twice.
3. In these plants, chloroplasts of mesophyll cells contain enzyme PEP carboxylase which fixes atmospheric C0₂. Thus, first C0₂ fixation occurs in mesophyll cells.
4. Decarboxylation of malic acid in bundle sheath cells results in increase in C0₂ concentration.
5. Thus, RuBisCO acts as carboxylase and brings about carboxylation of RuBP.
6. Due to this oxygenation of RuBP and photorespiration is prevented.
7. Thus, despite of having less number of bundle sheath cells carrying out Calvin cycle, C₄ plants are highly productive.

Question 22.
Who proposed C₄ pathway?
Answer:
1. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
2. It occurs in tropical and sub-tropical grasses and some dicotyledons.
3. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid. It is transported to the chloroplasts of bundle sheath cells.
3. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
4. Thus, concentration of C0₂ increases in the bundle sheath cells.
5. Chloroplasts of these cells contain enzymes of Calvin cycle.
6. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
7. Sugar formed in Calvin cycle is transported into the phloem.
8. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
9. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP. Thus, C₄ pathway needs 12 additional ATP.
10. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
11. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

Question 23.
Give schematic representation of HSK pathway.
Answer:
1. M. D. Hatch and C. R. Slack while working on sugarcane found four-carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
2. It occurs in tropical and sub-tropical grasses and some dicotyledons.
3. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.

Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid. It is transported to the chloroplasts of bundle sheath cells.
3. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm. Thus, concentration of C0₂ increases in the bundle sheath cells.
4. Chloroplasts of these cells contain enzymes of Calvin cycle.
5. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
6. Sugar formed in Calvin cycle is transported into the phloem.
7. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
8. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP. Thus, C₄ pathway needs 12 additional ATP.
9. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
10. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

Question 24.
With the help of labelled diagram explain Kranz anatomy.
Answer:

1. Anatomy of leaves of C₄ plants is different from leaves of C₃ plants.
2. C₄ plants show Kranz anatomy.
3. In the leaves of such plants, there is a bundle sheath around the vascular bundles.
4. The chloroplasts in the bundle – sheath cells are large and without or less developed grana, where as in the mesophyll cells the chloroplasts are small but with well-developed grana.

Question 25.
Give reason. Why C₄ plants are favoured in tropical regions?
Answer:

1. C₄ plants are favoured in tropical regions as they require 30 ATP to produce 1 molecule of glucose.
2. High temperature in tropical regions leads to closure of stomata to reduce rate of transpiration. Due to this availability of C0₂ decreases.
3. PEP carboxylase present in mesophyll cells can fix C0₂ even at low concentration. This helps the plant in efficient assimilation of atmospheric carbon dioxide.
4. C₄ plants contain a special leaf anatomy called Kranz anatomy which minimizes the losses due to photorespiration.
5. It helps C₄ plants to survive in conditions of high daytime temperatures, intense sunlight and low moisture.

Question 26.
With the help of suitable flowchart explain CAM.
Answer:
Crassulacean Acid Metabolism (CAM).

1. It is one more alternative pathway of carbon fixation found in desert plants.
2. It was first reported in the family Crassulaceae, therefore called as CAM (Crassulacean Acid Metabolism).
3. In CAM plants, stomata are scotoactive i.e. active during night, hence initial C0₂ fixation occurs in night.
4. Thus, C₄ pathway fix C0₂ at night and reduce C0₂ in day time via the C₃ pathway by using NADPH formed during the day.
5. PEP caboxylase and Rubisco are present in the mesophyll cell (no Kranz anatomy).
6. Formation of malic acid during dark is called acidification (phase I).
7. Malate is stored in vacuoles during the night.
8. Malate releases C0₂ during the day for C₃ pathway within the same cell is called deacidification (phase II).
9. Examples of CAM plants: Kalanchoe, Opuntia, Aloe etc.
10. The Chemical reactions of the carbon dioxide fixation and its assimilation are similar to that of C₄ plants.

Question 27.
Describe ‘any two’ factors affecting the rate of photosynthesis.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.

Internal factors which affects photosynthesis are as follows:
1. Chlorophyll:
a. Though presence of chlorophyll is essential for photosynthesis but rate of photosynthesis is proportional to the quantity of chlorophyll present.
b. It is because of the fact that chlorophyll merely acts as a biocatalyst and hence a small quantity is quite enough to maintain the large bulk of the reacting substances.
2. Sugar:
The final product in the photosynthesis reaction is sugar and its accumulation in the cells slow down the process of photosynthesis.
3. Internal structures:
The thickness of cuticle and epidermis of the leaf, the size and distribution of intercellular spaces and the distribution of the stomata and the development of chlorenchyma and other tissues also affects the rate of photosynthesis.

Question 28.
Enlist the factors that affect the rate of photosynthesis.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.

Internal factors which affects photosynthesis are as follows:
1. Chlorophyll:
a. Though presence of chlorophyll is essential for photosynthesis but rate of photosynthesis is proportional to the quantity of chlorophyll present.
b. It is because of the fact that chlorophyll merely acts as a biocatalyst and hence a small quantity is quite enough to maintain the large bulk of the reacting substances.
2. Sugar:
The final product in the photosynthesis reaction is sugar and its accumulation in the cells slow down the process of photosynthesis.
3. Internal structures:
The thickness of cuticle and epidermis of the leaf, the size and distribution of intercellular spaces and the distribution of the stomata and the development of chlorenchyma and other tissues also affects the rate of photosynthesis.

Question 29.
Give the Blackman’s law of limiting factors.
Answer:

1. The Blackman’s law of limiting factors states that when a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is controlled by the pace of the “slowest factor”.
2. The slowest factor is that factor which is present in the lowest or minimum concentration in relation to others.
3. The law of limiting factor can be explained by taking two external factors such as carbon dioxide and light.
4. For example, a plant photosynthesizing at a fixed light intensity sufficient to utilize 10mg of C₂ per hour only.
5. Photosynthetic rate goes on increasing when concentration of CO₂ increases.
6. Further increase in CO₂ concentration will not increase the rate of photosynthesis. In this case light becomes the limiting factor. Therefore, under such circumstances rate of photosynthesis can be increased only by increasing the light intensity.
7. This proves that the rate of photosynthesis responds to one factor alone at a time and there would be a sharp break in the curve and a plateau formed exactly at the point where another factor becomes limiting.
8. If any one of the other factors which is kept constant (e.g. Light) is increased, the photosynthetic rate increases again reaching the optimum where again another factor becomes limiting.

Question 44.
Multiple Choice Questions:

Question 1.
How many ATP molecules are required for synthesis of one glucose molecule using C₄ pathway?
(A) 30
(B) 18
(C) 6
(D) 2
Answer:
(A) 30

Question 2.
Photosynthesis is ________ reaction.
(A) oxidation
(B) reduction
(C) redox
(D) electrochemical
Answer:
(C) redox

Question 3.
Photosynthesis is minimum in _______ light.
(A) green
(B) blue
(C) red
(D) yellow
Answer:
(A) green

Question 4.
From the visible spectrum of light, which component is reflected by the green leaves?
(A) Blue
(B) Red
(C) Green
(D) Orange
Answer:
(C) Green

Question 5.
Which of the following is not an accessory pigment?
(A) Chlorophyll-b
(B) Xanthophyll
(C) Chlorophyll-a
(D) Carotene
Answer:
(C) Chlorophyll-a

Question 6.
The reaction centre of PS-II is ________ .
(A) Chi-a, 700
(B) Chi-a, 680
(C) Chi- a, 673
(D) Chi-a, 650
Answer:
(B) Chi-a, 680

Question 7.
Light reactions occur in ______________ .
(A) stroma
(B) grana
(C) matrix
(D) fret
Answer:
(B) grana

Question 8.
Dark reaction takes place in _____________ .
(A) Stroma
(B) Grana
(C) Matrix
(D) Thylakoid
Answer:
(A) Stroma

Question 9.
In dark reaction, the first compound to accept C0₂ is ___________ .
(A) RUMP
(B) RUBP
(C) PGAL
(D) PGA
Answer:
(B) RUBP

Question 10.
Which of the following is a photochemical reaction?
(A) Light reaction
(B) C₃ pathway
(C) C₄ pathway
(D) CAM pathway
Answer:
(A) Light reaction

Question 11.
Which of the following is a biochemical reaction?
(A) Light reaction
(B) Cyclic electron transfer
(C) Photolysis of water
(D) Dark phase
Answer:
(D) Dark phase

Question 12.
How many Calvin cycles are required to produce one molecule of glucose?
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(D) 6

Question 13.
The first C0₂ acceptor in C₄ pathway is _______ .
(A) Pyruvic acid
(B) PEPA
(C) OAA
(D) Malic acid
Answer:
(B) PEPA

Question 14.
In C₄ pathway, fixation of metabolic C0₂ occurs in which of the following cells?
(A) Bundle sheath cells
(B) Mesophyll cells
(C) Epidermal cells
(D) Cortical cells
Answer:
(A) Bundle sheath cells

Question 15.
Which one of the following is C₄ plant?
(A) Sunflower
(B) Soyabean
(C) Sugarcane
(D) Spinach
Answer:
(C) Sugarcane

Question 16.
Due to photorespiration, approximately __________ of photosynthetically fixed C0₂ is lost.
(A) 25%
(B) 50%
(C) 60%
(D) 80%
Answer:
(A) 25%

Question 17.
CAM plants are mostly ________________ .
(A) Tropical plants
(B) Succulents
(C) Monocots
(D) Mangroves
Answer:
(B) Succulents

Question 18.
Which of the following factors is not limiting?
(A) C0₂ concentration
(B) Light intensity
(C) Temperature
(D) Oxygen
Answer:
(D) Oxygen

Question 45.
Competitive Corner

Question 1.
In Hatch and Slack pathway, the primary C02 acceptor is _______________ .
(A) Rubisco
(B) Oxaloacetic acid
(C) Phosphoglyceric acid
(D) Phosphoenol pyruvate
Answer:
(D) Phosphoenol pyruvate

Question 2.
One scientist cultured Cladophora in a suspension of Azotobacter and illuminated the culture by splitting light through a prism. He observed that bacteria accumulated mainly in the region of:
(A) Blue and red light
(B) Violet and green light
(C) Indigo and green light
(D) Orange and yellow light
Hint: Cladophora is green alga and Azotobacter is aerobic bacteria. Theodor Engelmann split light into its spectral components using a prism and detected that aerobic bacteria accumulated mainly in the region of blue and red light.
Answer:
(A) Blue and red light

Question 3.
Clil-a and Chl-b shown maximum absorption in ________ regions of visible light.
(A) blue, violet and red
(B) red, indigo and green
(C) yellow, blue and red
(D) blue, violet and green
Answer:
(A) blue, violet and red

Question 4.
The co-enzyme which acts as hydrogen acceptor during light reaction is ________ .
(A) PQ
(B) FAD
(C) COQ
(D) NADP
Answer:
(D) NADP

Question 5.
During cyclic photophosphorylation, formation of ATP occurs between which of the following two compounds?
(A) FRS → Ferredoxin
(B) Cytochrome b₆ → Cytochrome f
(C) Cytochrome f → Plastocyanin
(D) Plastocyanin → Ionised Chi – a
Answer:
(B) Cytochrome b₆ → Cytochrome f

Question 6.
Which of the following is NOT a product of light reaction of photosynthesis?
(A) NADPH
(B) NADH
(C) ATP
(D) Oxygen
Answer:
(B) NADH

Question 7.
A part of photosynthetically fixed CO₂ goes back to the atmosphere due to _______.
(A) cyclic photophosphorylation
(B) noncyclic photophosphorylation
(C) dark reaction
(D) photorespiration
Answer:
(D) photorespiration

Question 8.
The highest rate of Photosynthesis in green plants are in _____ and ______ region of light spectrum.
(A) yellow and orange
(B) green and violet
(C) red and blue
(D) violet and blue
Answer:
(C) red and blue

Question 9.
Which one of the following is an essential factor for photophosphorylation?
(A) Sunlight
(B) Carbohydrate
(C) Oxygen
(D) Water
Answer:
(A) Sunlight

Question 10.
Cyclic photophosphorylation will NOT take place in the absence of ________ .
(A) carotenoids
(B) chlorophyll-a
(C) xanthophylls
(D) phycoerythrin
Answer:
(B) chlorophyll-a

Question 11.
Dark reaction of photosynthesis is a cyclic process as _______ is regenerated.
(A) RuBP
(B) C0₂
(C) Glucose
(D) PGA
Answer:
(A) RuBP

Question 12.
Phosphoenol pyruvate (PEP) is the primary CO₂ acceptor in:
(A) C₃ plants
(B) C₄ plants
(C) C₂ plants
(D) C₃ and C₄ plants
Answer:
(B) C₄ plants

Question 13.
In members of family Crassulaceae ________ is regenerated from starch during night.
(A) Phosphoenol pyruvic Acid
(B) Pyruvic Acid
(C) Malic Acid
(D) Oxalo Acetic Acid
Answer:
(A) Phosphoenol pyruvic Acid

Question 14.
With reference to factors affecting the rate of photosynthesis, which of the following statements are NOT correct?
(A) Light saturation for CO₂ fixation occurs at 10% of full sunlight
(B) Increasing atmospheric CO₂ concentration upto 0.05% can enhance C0₂ fixation rate
(C) C₃ plants responds to higher temperatures with enhanced photosynthesis while C₄ plants have much lower temperature optimum
(D) Tomato is a greenhouse crop which can be grown in CO₂ enriched atmosphere for higher yield.
Answer:
(C) C₃ plants responds to higher temperatures with enhanced photosynthesis while C₄ plants have much lower temperature optimum.

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 6 Redox Reactions Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 6 Redox Reactions

Question 1.
What does the term ‘redox’ refer to?
Answer:
Redox is an abbreviation used for the terms ‘oxidation and reduction’.

Question 2.
Give examples of naturally occurring redox reactions.
Answer:

1. Respiration
2. Rusting
3. Combustion of fuel

Question 3.
Define: Oxidant/Oxidising agent.
Answer:
A reagent/substance which itself undergoes reduction and causes oxidation of another species is called oxidant/oxidising agent.

Question 4.
Define: Reductant/Reducing agent
Answer:
A reagent/substance which itself undergoes oxidation bringing about the reduction of another species is called reductant/reducing agent.

Question 5.
Explain redox reaction giving an example.
Answer:
Oxidation and reduction reactions occur simultaneously. Therefore, oxidation-reduction reaction is also referred as redox reaction.

In the above reaction, HgCl₂ is reduced to Hg₂Cl₂ and SnCl₂ is oxidised to SnCl₄. Hence, it is a redox reaction.

Question 6.
Explain redox reaction in terms of electron transfer.
Answer:
i. Redox reaction can be described in terms of electron transfer as shown below:
2Mg_(s) + O_2(g) → Mg²⁺ + 2O^2-
ii. Charge development suggests that each magnesium atom loses two electrons to form Mg²⁺ and each oxygen atom gains two electrons to form O^2-. This can be represented as follows:

iii. When Mg is oxidised to MgO, the neutral Mg atom loses electrons to form Mg²⁺ in MgO while the elemental oxygen gains electrons and forms O^2- in MgO.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.

Question 7.
Justify the following reaction as redox reaction in terms of electron transfer.
Mg + F₂ → MgF₂
Answer:
i. Redox reaction can be described in terms of electron transfer as shown below:
Mg_(s) + F_2(g) → Mg²⁺ + 2F^–
ii. Charge development suggests that magnesium atom loses two electrons to form Mg²⁺ and each fluorine atom gains one electron to form F^–. This can be represented as follows:

iii. When Mg is oxidised to MgF₂, the neutral Mg atom loses electrons to form Mg²⁺ in MgF₂ while the elemental fluorine gains electrons and forms F^–in MgF₂.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.

Question 8.
Justify that the reaction 2Na_(s) + H_2(g) → 2NaH_(s) is a redox reaction.
Answer:
Redox reaction can be described as electron transfer as shown below:
2Na_(s) + H_2(g) → 2Na⁺ + 2H^–
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H^–. This can be represented as follows:

iii. When Na is oxidised to NaH, the neutral Na atom loses one electron to form Na⁺ in NaH while the elemental hydrogen gains one electron and forms H^– in NaH.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.

Question 9.
Define the terms oxidation and reduction in terms of electron transfer.
Answer:
i. The half reaction involving loss of electrons is called oxidation reaction.

ii. The half reaction involving gain of electrons is called reduction reaction.

Question 10.
Define the terms oxidant and reductant in terms of electron transfer.
Answer:

1. Oxidant: Oxidant or oxidising agent is an electron acceptor.
2. Reductant: Reductant or reducing agent is an electron donor.

Question 11.
Identify oxidising and reducing agents in the following reaction.
\(\mathrm{Fe}_{(s)}+\mathrm{Cu}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Fe}_{(\mathrm{aq})}^{2+}+\mathrm{Cu}_{(\mathrm{s})}\)
Answer:
Fe_(s) acts as a reducing agent as it donates electrons while \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) acts as an oxidising agent as it accepts electrons.

Question 12.
Define: Displacement reaction.
Answer:
A reaction in which an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element is called displacement reaction.
e.g. X + YZ → XZ + Y

Question 13.
Draw structure and assign oxidation number to each atom in:
i. Br₃O₈
ii. C₃O₂
Answer:
i. Br₃O₈

ii. C₃O₂

Question 14.
Deduce the oxidation number of S in the following species:
i. SO₂
ii. \(\mathrm{SO}_{4}^{2-}\)
Answer:
i. SO₂ is a neutral molecule.
∴ Sum of oxidation numbers of all atoms of SO₂ = 0
∴ (Oxidation number of S) + 2 × (Oxidation number of O) = 0
∴ Oxidation number of S + 2 × (- 2) = 0
∴ Oxidation number of S in SO₂ = 0 – (- 4)
∴ Oxidation number of S in SO₂ = +4

ii. \(\mathrm{SO}_{4}^{2-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms of \(\mathrm{SO}_{4}^{2-}\) = – 2
∴ (Oxidation number of S) + 4 × (Oxidation number of O) = – 2
∴ Oxidation number of S in \(\mathrm{SO}_{4}^{2-}\) = – 2 – 4 × (-2) = – 2 + 8
∴ Oxidation number of S in \(\mathrm{SO}_{4}^{2-}\) = +6

Question 15.
Assign oxidation number to each element in the following compounds or ions.
i. KMnO₄
ii. K₂Cr₂O₇
iii. Ca₃(PO₄)₂
Answer:
i. KMnO₄
Oxidation number of K = +1
Oxidation number of O = -2
KMnO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of K) + (Oxidation number of Mn) + 4 × (Oxidation number of O) = 0
∴ (+1) + Oxidation number of Mn + 4 × (-2) = 0
∴ Oxidation number of Mn + 1 – 8 = 0
∴ Oxidation number of Mn – 7 = 0
∴ Oxidation number of Mn in KMnO₄ = +7

ii. K₂Cr₂O₇
Oxidation number of K = +1
Oxidation number of O = -2
K₂Cr₂O₇ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of Cr) + 7 × (-2) = 0
∴ 2 × (Oxidation number of Cr) + 2 – 14 = 0
∴ 2 × (Oxidation number of Cr) – 12 = 0
∴ 2 × (Oxidation number of Cr) = +12
∴ Oxidation number of Cr = +12/2
∴ Oxidation number of Cr in K₂Cr₂O₇ = +6

iii. Ca₃(PO₄)₂
Oxidation number of Ca = +2 (∵ Ca is alkaline earth metal.)
Oxidation number of O = -2
Ca₃(PO₄)₂ is a neutral molecule.
Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of Ca) + 2 × (Oxidation number of P) + 8 × (Oxidation number of O) = 0
∴ 3 × (+2)+ 2 × (Oxidation number of P)+ 8 × (-2) = 0
∴ 2 × (Oxidation number of P) + 6 – 16 = 0
∴ 2 × (Oxidation number of P) – 10 = 0
∴ 2 × (Oxidation number of P) = +10
∴ Oxidation number of P = +10/2
∴ Oxidation number of P in Ca₃(PO₄)₂ = +5

Question 16.
Assign oxidation number to the atoms other than O and H in the following species.
i. \(\mathrm{SO}_{3}^{2-}\)
ii. \(\mathrm{BrO}_{3}^{-}\)
iii. \(\mathrm{ClO}_{4}^{-}\)
iv. \(\mathrm{NH}_{4}^{+}\)
v. \(\mathrm{NO}_{3}^{-}\)
vi. \(\mathrm{NO}_{2}^{-}\)
vii. SO₃
viii. N₂O₅
Answer:
The oxidation number of O atom bonded to a more electropositive atom is -2 and that of H atom bonded to electronegative atom is +1. Sum of the oxidation numbers of all atoms in ionic species is equal to charge it carries and that for neutral molecule is zero. Using these values, the oxidation numbers of atoms of the other elements in a given polyatomic species are calculated as follows:
i. \(\mathrm{SO}_{3}^{2-}\)
(Oxidation number of S) + 3 × (Oxidation number of O) = – 2
∴ Oxidation number of S + 3 × (-2) = – 2
∴ Oxidation number of S – 6 = – 2
∴ Oxidation number of S = – 2 + 6
∴ Oxidation number of S in \(\mathrm{SO}_{3}^{2-}\) = +4

ii. \(\mathrm{BrO}_{3}^{-}\)
(Oxidation number of Br) + 3 × (Oxidation number of O) = -1
∴ Oxidation number of Br + 3 × (-2) = – 1
∴ Oxidation number of Br – 6 = – 1
∴ Oxidation number of Br = – 1 + 6
Oxidation number of Br in \(\mathrm{BrO}_{3}^{-}\) = +5

iii. \(\mathrm{ClO}_{4}^{-}\)
(Oxidation number of Cl) + 4 × (Oxidation number of O) = – 1
∴ Oxidation number of Cl + 4 × (-2) = – 1
∴ Oxidation number of Cl – 8 = – 1
∴ Oxidation number of Cl = – 1 + 8
∴ Oxidation number of Cl in \(\mathrm{ClO}_{4}^{-}\) = +7

iv. \(\mathrm{NH}_{4}^{+}\)
(Oxidation number of N) + 4 × (Oxidation number of H) = + 1
∴ Oxidation number of N + 4 × (+1) = +1
∴ Oxidation number of N + 4 = + 1
∴ Oxidation number of N = + 1 – 4
∴ Oxidation number of N in \(\mathrm{NH}_{4}^{+}\) = -3

v. \(\mathrm{NO}_{3}^{-}\)
(Oxidation number of N) + 3 × (Oxidation number of O) = – 1
∴ Oxidation number of N + 3 × (-2) = – 1
∴ Oxidation number of N – 6 = – 1
∴ Oxidation number of N = – 1 + 6
∴ Oxidation number of N in \(\mathrm{NO}_{3}^{-}\) = +5

vi. \(\mathrm{NO}_{2}^{-}\)
(Oxidation number of N) + 2 × (Oxidation number of O) = – 1
∴ Oxidation number of N + 2 × (-2) = – 1
∴ Oxidation number of N – 4 = – 1
∴ Oxidation number of N = – 1 + 4
∴ Oxidation number of N in \(\mathrm{NO}_{2}^{-}\) = +3

vii. SO₃
(Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ Oxidation number of S + 3 × (-2) = 0
∴ Oxidation number of S – 6 = 0
∴ Oxidation number of S = 0 + 6
∴ Oxidation number of S in SO₃ = +6

viii. N₂O₅
2 × (Oxidation number of N) + 5 × (Oxidation number of O) = 0
∴ 2 × (Oxidation number of N) + 5 × (-2) = 0
∴ 2 × (Oxidation number of N) – 10 = 0
∴ 2 × (Oxidation number of N) = 0 + 10
∴ Oxidation number of N = 10/2
∴ Oxidation number of N in N₂O₅ = +5

Question 17.
Find the oxidation numbers of the underlined species in the following compounds or ions.

Answer:
i. P\(\mathrm{F}_{6}^{-}\)
Oxidation number of F = -1
P\(\mathrm{F}_{6}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ (Oxidation number of P) + 6 × (Oxidation number of F) = – 1
∴ Oxidation number of P + 6 × (-1) = -1
∴ Oxidation number of P – 6 = – 1
Oxidation number of P in P\(\mathrm{F}_{6}^{-}\) = +5

ii. NaIO₃
Oxidation number of Na = +1
Oxidation number of O = -2
NaIO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + (Oxidation number of I) + 3 × (Oxidation number of O) = 0
(+1) + (Oxidation number of I) + 3 × (-2) = 0
Oxidation number of I + 1 – 6 = 0
Oxidation number of I in NaIO₃ = +5

iii. NaHCO₃
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaHCO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of Na) + (Oxidation number of H) + (Oxidation number of C) + 3 × (Oxidation number of O) = 0
∴ (+1) + (+1) + (Oxidation number of C) + 3 × (-2) = 0
∴ Oxidation number of C + 2 – 6 = 0
∴ Oxidation number of C in NaHCO₃ = +4

iv. ClF₃
Oxidation number of F = -1
ClF₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of Cl) + 3 × (Oxidation number of F) = 0
∴ Oxidation number of Cl + 3 × (-1) = 0
∴ Oxidation number of Cl in ClF₃ = +3

v. Sb\(\mathrm{F}_{6}^{-}\)
Oxidation number of F = -1
Sb\(\mathrm{F}_{6}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ (Oxidation number of Sb) + 6 × (Oxidation number of F) = – 1
∴ Oxidation number of Sb + 6 × (-1) = -1
∴ Oxidation number of Sb in Sb\(\mathrm{F}_{6}^{-}\) = +5

vi. NaBH₄
Oxidation number of Na =+1
Oxidation number of H = -1 (for Hydride)
NaBH₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of Na) + (Oxidation number of B) + 4 × (Oxidation number of H) = 0
∴ (+1) + (Oxidation number of B) + 4 × (-1) = 0
∴ Oxidation number of B + 1 – 4 = 0
∴ Oxidation number of B in NaBH₄ = +3

vii. H₂PtCl₆
Oxidation number of H = +1
Oxidation number of Cl = -1
H₂PtCl₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of Pt) + 6 × (Oxidation number of Cl) = 0
∴ 2 × (+1) + (Oxidation number of Pt) + 6 × (-1) = 0
(Oxidation number of Pt) + 2 – 6 = 0
∴ Oxidation number of Pt in H₂PtCl₆ = +4

viii. H₅P₃O₁₀
Oxidation number of H = +1
Oxidation number of O = -2
H₅P₃O₁₀ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 5 × (Oxidation number of H) + 3 × (Oxidation number of P) +10 × (Oxidation number of O) = 0
∴ 5 × (+1) + 3 × (Oxidation number of P) + 10 × (-2) = 0
∴ 3 × (Oxidation number of P) + 5 – 20 = 0
Oxidation number of P = +\(\frac {15}{3}\)
∴ Oxidation number of P in H₅P₃O₁₀ = +5

ix. V₂\(\mathrm{O}_{7}^{4-}\)
Oxidation number of O = -2
V₂\(\mathrm{O}_{7}^{4-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 4
∴ 2 × (Oxidation number of V) + 7 × (Oxidation number of O) = – 4
∴ 2 × (Oxidation number of V) + 7 × (-2) = – 4
∴ 2 × (Oxidation number of V) = – 4 + 14
∴ Oxidation number of V = +\(\frac {10}{2}\)
∴ Oxidation number of V in V₂\(\mathrm{O}_{7}^{4-}\) = +5

x. CuSO₄
Oxidation number of Cu = +2
Oxidation number of O = -2
CuSO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of Cu) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ (+2) + Oxidation number of S + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in CuSO₄ = +6

xi. Bi\(\mathrm{O}_{3}^{-}\)
Oxidation number of O = -2
Bi\(\mathrm{O}_{3}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ (Oxidation number of Bi) + 3 × (Oxidation number of O) = – 1
∴ Oxidation number of Bi + 3 × (-2) = – 1
∴ Oxidation number of Bi = – 1 + 6
∴ Oxidation number of Bi in Bi\(\mathrm{O}_{3}^{-}\) = +5

xii. CH₃OH
Oxidation number of H = +1
Oxidation number of O = -2
CH₃OH is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of C) + 4 × (Oxidation number of H) + (Oxidation number of O) = 0
∴ (Oxidation number of C) + 4 × (+1) + (-2) = 0
∴ Oxidation number of C + 2 = 0
∴ Oxidation number of C in CH₃OH = -2

xiii. H₂O₂
Oxidation number of O = -1 (for peroxide)
H₂O₂ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 2 × (Oxidation number of O) = 0
∴ 2 × (Oxidation number of H) + 2 × (-1) = 0
∴ Oxidation number of H = +\(\frac {2}{2}\)
∴ Oxidation number of H in H₂O₂ = +1

xiv. C₄H₄\(\mathrm{O}_{6}^{2-}\)
Oxidation number of H = +1
Oxidation number of O = -2
C₄H₄\(\mathrm{O}_{6}^{2-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 4 × (Oxidation number of C) + 4 × (Oxidation number of H) + 6 × (Oxidation number of O) = – 2
∴ 4 × (Oxidationnumber of C) + 4 × (+1) +6 × (-2) = -2
∴ 4 × (Oxidation number of C) + 4 – 12 = -2
∴ 4 × (Oxidation number of C) = – 2 + 8
∴ Oxidation number of C = +\(\frac {6}{4}\)
∴ Oxidation number of C in C₄H₄\(\mathrm{O}_{6}^{2-}\) = +1.5

xv. H₂As\(\mathrm{O}_{4}^{-}\)
Oxidation number of H = +1
Oxidation number of O = -2
H₂As\(\mathrm{O}_{4}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ 2 × (Oxidation number of H) + (Oxidation number of As) + 4 × (Oxidation number of O) = -1
∴ 2 × (+1) + Oxidation number of As + 4 × (-2) = – 1
∴ Oxidation number of As + 2 – 8 = – 1
∴ Oxidation number of As = – 1 + 6
∴ Oxidation number of As in H₂As\(\mathrm{O}_{4}^{-}\) = +5

xvi. Mn(OH)₃
Oxidation number of O = -2
Oxidation number of H = +1
Mn(OH)₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ (Oxidation number of Mn) + 3 × (Oxidation number of O) + 3 × (Oxidation number of H) = 0
∴ Oxidation number of Mn + 3 × (-2) + 3 × (+1) = 0
∴ Oxidation number of Mn – 6 + 3 = 0
∴ Oxidation number of Mn in Mn(OH)₃ = +3

xvii. \(\mathrm{I}_{3}^{-}\)
\(\mathrm{I}_{3}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ 3 × Oxidation number of I = – 1
∴ Oxidation number of I in \(\mathrm{I}_{3}^{-}\) = –\(\frac {1}{3}\)

xviii. C₂H₅OH
Oxidation number of O = -2
Oxidation number of H = +1
C₂H₅OH is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of C) + 6 × (Oxidation number of H) + (Oxidation number of O) = 0
∴ 2 × (Oxidationnumberof C) + 6 × (+1) + (-2) = 0
∴ 2 × (Oxidation number of C) = – 4
∴ Oxidation number of C = –\(\frac {4}{2}\)
∴ Oxidation number of C in C₂H₅OH = -2

xix. Na₂CO₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂CO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + (Oxidation number of C) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of C) + 3 × (-2) = 0
∴ Oxidation number of C + 2 – 6 = 0
∴ Oxidation number of C in Na₂CO₃ = +4

xx. I[latex]\mathrm{O}_{4}^{-}[/latex]
Oxidation number of O = -2
I\(\mathrm{O}_{4}^{-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ (Oxidation number of I) + 4 × (Oxidation number of O) = – 1
∴ Oxidation number of I + 4 × (-2) = – 1
∴ Oxidation number of I = -1 +8
∴ Oxidation number of I in I\(\mathrm{O}_{4}^{-}\) = +7

xxi. V\(\mathrm{O}_{4}^{3-}\)
Oxidation number of O = -2
V\(\mathrm{O}_{4}^{3-}\) is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 3
∴ (Oxidation number of V) + 4 × (Oxidation number of O) = – 3
∴ Oxidation number of V + 4 × (-2) = – 3
∴ Oxidation number of V = -3 + 8
∴ Oxidation number of V in V\(\mathrm{O}_{4}^{3-}\) = +5

xxii. Ni₂O₃
Oxidation number of O = -2
Ni₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Ni) + 3 × (Oxidation number of O) = 0
∴ 2 × (Oxidation number of Ni) + 3 × (-2) = 0
∴ 2 × (Oxidation number of Ni) = +6
∴ Oxidation number of Ni = +\(\frac {6}{2}\)
∴ Oxidation number of Ni in Ni₂O₃ = +3

xxiii. K₃[Fe(CN)₆]
Oxidation number of K = +1
Oxidation number of CN group = -1
K₃[Fe(CN)₆] is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of K) + (Oxidation number of Fe) + 6 × (Oxidation number of CN group) = 0
∴ 3 × (+1) + Oxidation number of Fe + 6 × (-1) = O
∴ Oxidation number of Fe + 3 – 6 = 0
∴ Oxidation number of Fe in K₃[Fe(CN)₆] = +3

Question 18.
Define: Stock notation.
Answer:
Representation in which oxidation number of an atom is denoted by Roman numeral in parentheses after the chemical symbol is called Stock notation. This name was given after the German scientist, Alfred Stock. e.g. Au¹⁺Cl^1- → Au(I)Cl

Question 19.
What is the use of Stock notation?
Answer:
The Stock notation is used to specify the oxidation number of the metal.

Question 20.
How will you write Stock notations for the following compounds?
i. AuCl₃
ii. SnCl₄
iii. SnCl₂
iv. MnO₂
Answer:
i. AuCl₃: The charge on each element is \(\mathrm{Au}^{3+} \mathrm{Cl}_{3}^{1-}\). Hence, the stock notation is Au(III)Cl₃.
ii. SnCl₄: The charge on each element is \(\mathrm{Sn}^{4+} \mathrm{Cl}_{4}^{1-}\). Hence, the stock notation is Sn(IV)Cl₄.
iii. SnCl₂: The charge on each element is \(\mathrm{Sn}^{2+} \mathrm{Cl}_{2}^{1-}\). Hence, the stock notation is Sn(II)Cl₂.
iv. MnO₂: The charge on each element is \(\mathrm{Mn}^{4+} \mathrm{O}_{2}^{2-}\). Hence, the stock notation is Mn(IV)O₂.

Question 21.
Write the formula for each of the following ionic compounds:
i. Nickel(III) oxide
ii. Tin(IV) chloride
iii. Bismuth(V) chloride
iv. Cobalt(III) chloride
v. Lead(IV) oxide
vi. Chromium(II) chloride
Answer:
i. Ni₂O₃
ii. SnCl₄
iii. BiCl₅
iv. CoCl₃
v. PbO₂
vi. CrCl₂

Question 22.
Define the terms oxidation and reduction in terms of oxidation number.
Answer:
i. Oxidation is an increase in the oxidation number of an element in a given substance.
e.g. Fe_(s) → \(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\)
ii. Reduction is a decrease in the oxidation number of an element in a given substance. e.g. \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) → Cu_(s)

Question 23.
Define the terms oxidant and reductant in terms of oxidation number.
Answer:

1. Oxidant: Oxidant or oxidising agent is a substance which increases the oxidation number of an element in a given substance, and itself undergoes decrease in oxidation number of a constituent element in it.
2. Reductant: Reductant or reducing agent is a substance that lowers the oxidation number of an element in a given substance, and itself undergoes an increase in the oxidation number of a constituent element in it.

Question 24.
Identify whether the following reaction is redox or NOT. State oxidant and reductant therein.

Answer:
i. Write oxidation number of all the atoms of reactants and products.

ii. Identify the species that undergoes change in oxidation number.

iii. The oxidation number of As increases from +3 to +5 and that of Br decreases from +5 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
iv. The oxidation number of As increases by loss of electrons and therefore, As is a reducing agent and itself is oxidised. On the other hand, the oxidation number of Br decreases and therefore, Br is an oxidising agent and itself is reduced by gain of electrons.
Result:
a. The given reaction is a redox reaction.
b. Oxidant/oxidising agent: \(\mathrm{BrO}_{3}^{-}\)
c. Reductant/reducing agent: H3AsO3

Question 25.
For the reaction, \(\mathrm{SeO}_{3(\mathrm{aq})}^{2-}+\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{SeO}_{4(\mathrm{aq})}^{2-}+2 \mathrm{Cl}_{(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\), complete the following table.

Oxidising agent	————–
Reducing agent	————–
Oxidised species	————–
Reduced species	————–

Answer:

Oxidising agent	Cl2(g)
Reducing agent	\(\mathrm{Se} \mathrm{O}_{3(\mathrm{aq})}^{2-}\)
Oxidised species	\(\mathrm{Se} \mathrm{O}_{4(\mathrm{aq})}^{2-}\)
Reduced species	\(2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Question 26.
Using oxidation number concept, identify the redox reactions, identify oxidizing and reducing agents in case of redox reactions.

Answer:
i. H₃PO_4(aq) + 3KOH_(aq) → K₃PO_4(aq) + 3H₂O_(l)
a. Write oxidation number of all the atoms of reactants and products

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction.

Result:
The given reaction is NOT a redox reaction.

ii. Zn_(s) + 2HCl_(aq) → ZnCl₂(_(aq)) + H_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of Zn increases from 0 to +2 and that of H decreases from +1 to 0. Because oxidation number of one species increases and that of other decreases, the reaction is redox reaction.
d. The oxidation number of Zn increases by loss of electrons and therefore, Zn is a reducing agent and itself is oxidized. On the other hand, the oxidation number of H decreases by gain of electrons and therefore, H is oxidising an agent and itself is reduced by gain of electrons.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent: HCl
3. Reductant/reducing agent: Zn

iii.

c. The oxidation number of Fe increases from +2 to +3 and that of Br decreases from +5 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of Fe increases by loss of electrons and therefore, Fe is a reducing agent and itself is oxidized. On the other hand, the oxidation number of Br decreases by gain of electrons and therefore, Br is an oxidising an agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent: \(\mathrm{BrO}_{3}^{-}\)
3. Reductant/reducing agent: Fe²⁺

iv. 2Zn_(s) + O_2(g) → 2ZnO_(s)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of Zn increases from 0 to +2 and that of O decreases from 0 to -2. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of Zn increases by loss of electrons and therefore, Zn is a reducing agent and itself is oxidized. On the other hand, the oxidation number of O decreases by gain of electrons and therefore, O is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent: O₂
3. Reductant/reducing agent: Zn

v. \(\mathrm{Sn}_{(a q)}^{2+}+\mathrm{IO}_{4(2 q)}^{-} \longrightarrow \mathrm{Sn}_{(aq)}^{4+}+\mathrm{I}_{(a \mathrm{q})}^{-}\)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of Sn increases from +2 to +4 and that of I decreases from +7 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of Sn increases by loss of electrons and therefore, Sn is a reducing agent and itself is oxidized. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent: Sn²⁺
3. Reductant/reducing agent: \(\mathrm{IO}_{4}^{-}\)

Question 27.
Name two methods used to balance redox reactions.
Answer:

1. Oxidation number method
2. Half reaction method or ion electrode method

Question 28.
Describe the steps involved in balancing redox reactions by the oxidation number method.
Answer:
i. Step 1: Write the unbalanced equation for redox reaction. Balance the equation for all atoms in the reactions, except H and O. Identify the atoms which undergo change in oxidation number and by how much. Draw the bracket to connect atoms of the elements that changes the oxidation number.

ii. Step 2: Show an increase in oxidation number per atom of the oxidised species and hence, the net increase in oxidation number. Similarly, show a decrease in the oxidation number per atom of the reduced species and the net decrease in oxidation number. Determine the factors which will make the total increase and decrease equal. Insert the coefficients into the equation.

iii. Step 3: Balance oxygen atoms by adding H₂O to the side containing less O atoms, one H₂O is added for one O atom. Balance H atoms by adding H⁺ ions to the side having less H atoms.

iv. Step 4: If the reaction occurs in basic medium, then add OH^– ions equal to the number of H⁺ ions added in step 3, on both the sides of equation. The H⁺ and OH^– ions on same side of reactions are combined to give H₂O molecules.

v. Step 5: Check the equation with respect to both, the number of atoms of each element and the charges. It is balanced.
Note: For acidic medium, step 4 is omitted.

Question 29.
Using the oxidation number method write the net ionic equation for the reaction of potassium permanganate, KMnO₄, with ferrous sulphate, FeSO₄.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Fe}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+\mathrm{Fe}_{(\mathrm{aq})}^{3+}\)
Answer:
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Fe}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+\mathrm{Fe}_{(\mathrm{aq})}^{3+}\)

Step 2: Assign oxidation number to Mn and Fe, and calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and dicrease equal, we must take 5 atoms of Fe²⁺
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+5 \mathrm{Fe}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+5 \mathrm{Fe}_{(\mathrm{aq})}^{3+}\)

Step 3: Balance the ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+5 \mathrm{Fe}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+5 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H⁺ on the left-hand side.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+5 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+5 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 5: Check the two sides for balance of charges and atoms. The net ionic equation obtained in step 4 is the balanced equation.
Hence, balanced equation:
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+5 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Mn}_{(\mathrm{aq})}^{2+}+5 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Question 30.
Balance the following reaction by oxidation number method.
CuO + NH₃ → Cu + N₂ + H₂O
Answer:
Step 1: Write skeletal equation and balance the elements other than O and H.
CuO + 2NH₃ → Cu + N₂ + H₂O
Step 2: Assign oxidation number to Cu and N. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Cu and 2 atoms of N. (There are already 2 N atoms.)
3CuO + 2NH₃ → 3Cu + N₂ + H₂O
Step 3: Balance ‘O’ atoms by adding 3H₂O to the right-hand side.
3CuO + 2NH₃ → 3Cu + N₂ + 3H₂O
Step 4: Charges are already balanced.
Step 5: Check two sides for balance of atoms and charges. The equation obtained in step 3 is balanced.
Hence, balanced equation: 3CuO + 2NH₃ → 3Cu + N₂ + 3H₂O

Question 31.
Balance the following redox equation by oxidation number method. The reactions occur in acidic medium.

Answer:
i. \(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7 \text { (aq) }}^{2-} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+\mathrm{Cr}_{(\mathrm{aq})}^{3+}\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7 \text { (aq) }}^{2-} \longrightarrow \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}\)

Step 2: Assign oxidation number to O and Cr. Calculate the increase and decrease in the oxidation number and make them equal.

Since there are two Cr atoms, the net decrease in oxidation number is 6. In order to make the net increase and decrease equal, we must take 6 atoms of O i.e., 3H₂O₂.
\(3 \mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-} \longrightarrow 3 \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}\)

Step 3: Balance ‘O’ atoms by adding 7H₂O to the right-hand side.
\(3 \mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-} \longrightarrow 3 \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+7 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H⁺ on the left-hand side.
\(3 \mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+7 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:
\(3 \mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 3 \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+7 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

ii. \(\mathrm{Ag}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{q})}^{-} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{2(\mathrm{~g})}\)

Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Ag}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{q})}^{-} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{2(\mathrm{~g})}\)

Step 2: Assign oxidation number to Ag and N. Calculate the increase and decrease in the oxidation number and make them equal.

Because net increase is equal to net decrease, multiplying coefficients are not required.

Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(\mathrm{Ag}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.
\(\mathrm{Ag}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Thus, the equation is balanced with respect to the atoms as well as charges.

Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \(\mathrm{Ag}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)

iii. \(\mathrm{Sn}_{(\mathfrak{a q})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-} \longrightarrow \mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{I}_{(\mathrm{aq})}^{-}\)

Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Sn}_{(\mathfrak{a q})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-} \longrightarrow \mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{I}_{(\mathrm{aq})}^{-}\)

Step 2: Assign oxidation number to Sn and I. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 4 atoms of Sn.
\(4 \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-} \longrightarrow 4 \mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{I}_{(\mathrm{aq})}^{-}\)

Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(4 \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-} \longrightarrow 4 \mathrm{Sn}_{(\text {aq })}^{4+}+\mathrm{I}_{(\text {aq })}^{-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H⁺ on the left-hand side.
\(4 \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 4 \mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{I}_{(\mathrm{aq})}^{-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \(4 \mathrm{Sn}_{(\mathrm{aq})}^{2+}+\mathrm{IO}_{4(\mathrm{aq})}^{-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 4 \mathrm{Sn}_{(\mathrm{aq})}^{4+}+\mathrm{I}_{(\mathrm{aq})}^{-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

iv. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{IO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{IO}_{4(\mathrm{aq})}^{-}\)

Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{IO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{IO}_{4(\mathrm{aq})}^{-}\)

Step 2: Assign oxidation number to Mn and I. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of Mn and 3 atoms of I.
\(2 \mathrm{MnO}_{4(a q)}^{-}+3 \mathrm{IO}_{3(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+3 \mathrm{IO}_{4(\mathrm{aq})}^{-}\)

Step 3: Balance ‘O’ atoms by adding 1H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+3 \mathrm{IO}_{3(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+3 \mathrm{IO}_{4(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 2H on the left-hand side.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+3 \mathrm{IO}_{3(\mathrm{aq})}^{-}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+3 \mathrm{IO}_{4(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)

Step 5: Check two sides for balanced of atoms and charges.
Hence, balanced equation: \(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+3 \mathrm{IO}_{3(\mathrm{aq})}^{-}+2 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+3 \mathrm{IO}_{4(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)

Question 32.
Balance the following redox equation in basic medium by oxidation number method:

Answer:
i. \(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{NH}_{3(\mathrm{aq})}+\mathrm{Zn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)

Step 1: Write skeletal equation and balance the elements other than 0 and H.
\(\mathrm{Zn}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{NH}_{3(\mathrm{aq})}+\mathrm{Zn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)

Step 2: Assign oxidation number to Zn and N. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Zn.
\(2 \mathrm{Zn}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{NH}_{3(\mathrm{aq})}+2 \mathrm{Zn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)

Step 3: Balance ‘O’ atoms by adding 9H₂O to the left-hand side.
\(2 \mathrm{Zn}_{(\mathrm{s})}+\mathrm{NO}_{3(\mathrm{aq})}^{-}+9 \mathrm{H}_{2} \mathrm{O}_{(l)} \longrightarrow \mathrm{NH}_{3(\mathrm{aq})}+2 \mathrm{Zn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3H on the right-hand side.

Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{HO}_{2(\mathrm{aq})}^{-} \longrightarrow \mathrm{Mn}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnCl}_{2(\mathrm{aq})}+\mathrm{HO}_{2(\mathrm{aq})}^{-} \longrightarrow \mathrm{Mn}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Step 2: Assign oxidation number to Mn and O. Calculate the increase and decrease in the oxidation number and make them equal.

iii. Cu(OH)_2(s) + N₂H_4(aq) → Cu_(s) + N_2(g)
Step 1: Write skeletal equation and balance the elements other that O and H.
Cu(OH)_2(s) + N₂H_4(aq) → Cu_(s) + N_2(g)
Step 2: Assign oxidation number to Cu and N. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of Cu.
2Cu(OH)_2(s) + N₂H_4(aq) → 2Cu_(s) + N_2(g)
Step 3: Balance ‘O’ atoms by adding 4H20 to the right-hand side.
2Cu(OH)_2(s) + N₂H_4(aq) → 2Cu_(s) + N_2(g) + 4H₂O_(l)
Step 4: The equation is balanced for both atoms as well as charges.
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2Cu(OH)_2(s) + N₂H_4(aq) → 2Cu_(s) + N_2(g) + 4H₂O_(l)

Question 33.
Describe the steps involved in balancing redox reactions by ion electron method (Half reaction method).
Answer:
In this method two half equations are balanced separately and then added together to give balanced equation. Following steps are involved.

• Step 1: Write unbalanced equation for the redox reaction, assign oxidation number to all the atoms in the reactants and products. Divide the equation into two half equations. One half equation involves increase in oxidation number and another involves decrease in oxidation number (Write two half equations separately).
• Step 2: Balance the atoms except O and H in each half equation. Balance oxygen atom by adding H₂O to the side with less O atoms.
• Step 3: Balance H atoms by adding H⁺ ions to the side having less H atoms.
• Step 4: Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation and to the left of reduction half equation.
• Step 5: Multiply half equation by suitable factors to equalize the number of electrons in two half equations. Add two half equations and cancel the number of electrons on both sides of equation.
• Step 6: If the reaction occurs in basic medium then add OH^– ions, equal to number of H⁺ ions on both sides of equation. The H⁺ and OH^– ions on same side of equation combine to give H₂O molecules.
• Check that the equation is balanced in both, the atoms and the charges.

Question 34.
Balance the following unbalanced equation (in acidic medium) by ion electron method (half reaction method).
\(\mathrm{Mn}_{(\mathrm{aq})}^{2+}+\mathrm{ClO}_{3(\mathrm{qq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{ClO}_{2(\mathrm{aq})}\)
Answer:
Step 1: Write imbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 2H₂O to left side of oxidation half equation and 1H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H+ ions to the side with less H. Hence add 4H⁺ ions to the right side of oxidation half equation and 2H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 1 electron to the left side of reduction half equation to balance the charges.

Step 5: Multiply reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Add two half equations:
\(\mathrm{Mn}_{(\mathrm{aq})}^{2+}+2 \mathrm{ClO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+2 \mathrm{ClO}_{2(\mathrm{aq})}\)
The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation: \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}+2 \mathrm{ClO}_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+2 \mathrm{ClO}_{2(\mathrm{aq})}\)

Question 35.
Balance the following unbalanced equation by ion electron method (half reaction method).
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})}+\mathrm{ClO}_{4(\mathrm{aq})}^{-} \longrightarrow \mathrm{ClO}_{2(\mathrm{aq})}^{-}+\mathrm{O}_{2(\mathrm{~g})}\)
Answer:
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance the half equation for O atoms by adding H₂O to the side with less O atoms. Hence, add 2H₂O to the right side of reduction half equation and none to the oxidation half equation

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence add 2H⁺ ions to the right side of oxidation half equation and 4H⁺ ions to the left side of reduction half equation.

Step 4: Add 2 electrons to the right side of oxidation half equation and 4 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 2 to equalize the number of electrons and then add two half equations.

Question 36.
Balance the following redox equations by ion-electron (half reaction method).

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{Fe}_{(\mathrm{aq})}^{2+} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{Fe}_{(\mathrm{aq})}^{3+}(\text { acidic })\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 7H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H+ ions to the side with less H. Hence, add 14H⁺ ions to the left side of reduction half equation.

Step 4: Now add 1 electron to the right side of oxidation half equation and 6 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 6 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{SO}_{2(\mathrm{~g})}+\mathrm{Fe}_{(\mathrm{aq})}^{3+} \longrightarrow \mathrm{Fe}_{(\mathrm{aq})}^{2+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}(\text { acidic })\)
Step 1 : Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products in them. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 2H₂O to the left side of oxidation half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 4H⁺ ions to the right side of oxidation half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 1 electron to the left side of reduction half equation to balance the charges.

Step 5: Multiply reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

iii. \(\mathrm{ClO}_{(\mathrm{aq})}^{-}+\mathrm{Cr}(\mathrm{OH})_{4(\mathrm{aq})}^{-} \longrightarrow \mathrm{CrO}_{4(\mathrm{aq})}^{2-}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \text {(basic) }\)
Step 1: Write unbalanced equation for the redox reaction: Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 1H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 4H⁺ ions to the right side of oxidation half equation and 2H⁺ ions to the left side of reduction half equation.

Step 4: Now add 3 electrons to the right side of oxidation half equation and 2 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 2 and reduction half equation by 3 to equalize number of electrons in two half equations. Then add two half equation.

Step 6: The reaction takes place in basic medium. 20H^– ions, equal to the number of H⁺ ions (2H⁺ ions) are added on both sides of the equation.

iv. \(\mathrm{SeO}_{3(\mathrm{aq})}^{2-}+\mathrm{Cl}_{2(\mathrm{~g})} \longrightarrow \mathrm{SeO}_{4(a q)}^{2-}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \text {(basic) }\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products in them. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding H₂O to the left side of oxidation half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 2 electrons to the left side of reduction half equation to balance the charges.

Step 5: There is equal number of electrons in two half equations. Add two half equation.

Question 37.
Explain displacement reaction in terms of redox reaction by giving example.
Answer:
i. Displacement reaction can be looked upon as redox reaction. Consider the following displacement reaction.

ii. Here, Zn gets oxidized to Zn²⁺ ion and Cu²⁺ ions get reduced to metallic Cu. A direct transfer of electron from zinc atom to cupric ions takes place in this case.

Question 38.
Explain construction of Daniel cell.
Answer:

• The zinc and copper plates are connected by an electric wire through a switch and voltmeter.
• The solution in two containers are connected by salt bridge (U-shaped glass tube containing a gel of KCl or NH₄NO₃ in agar-agar).
• When switch is on, electrical circuit is complete as indicated by the deflection in the voltmeter.
• The circuit has two parts, one in the form of electrical wire which allows the flow of electrons and the other in the form of two solutions joined by salt bridge. In solution part of the circuit, the electric current is carried by movement of ions.

Question 39.
Explain working of Daniel cell.
Answer:
i. When a circuit is complete, the zinc atoms on zinc plates spontaneously lose electrons which are picked up in the external circuit.
ii. The electrons flow from the zinc plate to copper plate through wire.
iii. Cu²⁺ ions in the second container receive these electrons through the copper plate and are reduced to copper atoms which get deposited on the copper plate.
iv. Here, zinc plate acts as anode (negative electrode) and the copper plate acts as cathode (positive electrode).

v. Thus, when two half reactions, namely, oxidation and reduction, are allowed to take place in separate containers and provision is made for completing the electrical circuit, electron transfer take place through the circuit.
vi. This results in flow of electric current in the circuit as indicated by deflection in voltmeter.
vii. Thus, in Daniel cell, electricity is generated by redox reaction.

Question 40.
Draw a neat and labelled diagram of Daniel cell.
Answer:

Question 41.
Chalcopyrite (CuFeS₂) is a common ore of copper. Since it has low concentration of copper, the ore is first concentrated through froth floatation process. The concentrated ore is then heated strongly with silicon dioxide (silica) and oxygen in a furnace. The product obtained, copper(I) sulphide, is further converted to copper (99.5% pure) with a final blast of air (O₂) during which sulphur dioxide is released as a by-product.
i. Write a balanced reaction for the extraction of copper from copper(I) sulphide.
ii. Which species undergoes an increase in the oxidation state?
iii. Which species accepts electrons?
Answer:
i. Cu₂S + O₂ → 2Cu + SO₂
ii. Sulphur undergoes an increase in the oxidation state from -2 (in Cu₂S) to +4 (in SO₂).
iii. Copper accepts one electron and undergoes a decrease in the oxidation state from +1(in Cu₂S) to 0 (in Cu). Oxygen accepts two electrons and undergoes a decrease in the oxidation state from 0 (in O₂) to -2 (in SO₂).

Question 42.
Consider the elements: Cs, Ne, I and F
i. Identify the element that exhibits only negative oxidation state.
ii. Identify the element that exhibits only positive oxidation state.
iii. Identify the element that exhibits both negative as well as positive oxidation state.
iv. Identify the element that exhibits neither the negative nor the positive oxidation state.
Answer:
i. F: It is most electronegative. It shows only a negative oxidation state of -1.
ii. Cs: Alkali metals have only one electron in their valence shell and hence, exhibits only positive (+1) oxidation state.
iii. I: Because of the presence of 7 electrons in its valence shell, I shows negative oxidation state of -1 (to have stable noble gas configuration) and positive oxidation numbers of +1, +3, +5 and +7 because of the presence of d-orbitals.
iv. Ne: It is an inert gas and therefore, does not exhibit negative or positive oxidation state.

Multiple Choice Questions

1. Loss of electrons means ………….
(A) reduction
(B) oxidation
(C) precipitation
(D) complexometry
Answer:
(B) oxidation

2. Reduction involves ……………
(A) gain of electrons
(B) addition of oxygen
(C) increase in oxidation number
(D) loss of electron
Answer:
(A) gain of electrons

3. Which of the following statement is INCORRECT?
(A) Oxidant is a substance which increases the oxidation number of other substance.
(B) Reductant is a substance which decreases the oxidation number of other substance.
(C) The oxidation number of oxidant decreases.
(D) In oxidation, there is decrease in oxidation number.
Answer:
(D) In oxidation, there is decrease in oxidation number.

4. Which of the following is an example of oxidation process?

Answer:
(D) \(\mathrm{Li}_{(\mathrm{s})} \longrightarrow \mathrm{Li}_{(\mathrm{g})}^{+}+\mathrm{e}^{-}\)

5. Which is the best description of the behaviour of chlorine in the reaction?
H₂O + Cl₂ → HOCl + HCl
(A) Neither oxidized not reduced
(B) Both oxidised and reduced
(C) Oxidised only
(D) Reduced only
Answer:
(B) Both oxidised and reduced

6. In the reaction,
\(\begin{array}{r} 3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \ 5 \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}^{-} \end{array}\) …………..
(A) Br₂ is oxidised and carbonate is reduced
(B) bromine is reduced and water is oxidised
(C) bromine is neither reduced nor oxidised
(D) bromine is both reduced and oxidised
Answer:
(D) bromine is both reduced and oxidised

7. A chemical reaction in which oxidation and reduction processes takes place simultaneously is known as ………… reaction.
(A) redox
(B) precipitation
(C) complexometric
(D) titration
Answer:
(A) redox

8. Which of the following is a redox reaction?
(A) NaCl + KNO₃ → NaNO₃ + KCl
(B) CaC₂O₄ + 2HCl → CaCl₂ + H₂C₂O₄
(C) Mg(OH)₂ + 2NH₄Cl → MgCl₂ + 2NH₄OH
(D) Zn + 2AgCN → 2Ag + Zn(CN)₂
Answer:
(D) Zn + 2AgCN → 2Ag + Zn(CN)₂

9. Oxidation number of metal ion is always …………..
(A) positive
(B) negative
(C) zero
(D) non zero
Answer:
(A) positive

10. The oxidation number of oxygen in peroxide is …………..
(A) -2
(B) -1
(C) +1
(D) +2
Answer:
(B) -1

11. The oxidation number of oxygen is …………. in oxygen difluoride.
(A) -2
(B) -1
(C) +2
(D) +1
Answer:
(C) +2

12. Oxidation number of carbon in CH₂F₂ is ………….
(A) +1
(B) -1
(C) 0
(D) +2
Answer:
(C) 0

13. In calcium hydride (CaH₂), the oxidation number of hydrogen is ………….
(A) +1
(B) -1
(C) +2
(D) -2
Answer:
(B) -1

14. The element with atomic number 9 can exhibit oxidation state of …………..
(A) +1
(B) +3
(C) -1
(D) +5
Answer:
(C) -1

15. The highest and lowest oxidation states possible for Te (group 16) are ……………
(A) +6, -2
(B) +6, 0
(C) +4, -4
(D) +6, -6
Answer:
(A) +6, -2

16. What is the oxidation state of S in Na₂S₂ ?
(A) +1
(B) -2
(C) -1
(D) 0
Answer:
(C) -1

17. The oxidation state of S in S₂O₈^2- is ………….
(A) +2
(B) + 4
(C) +6
(D) + 7
Answer:
(D) + 7

18. The oxidation state of phosphorus in Ba(H₂PO₂)₂ is …………..
(A) +3
(B) +2
(C) +1
(D) -1
Answer:
(C) +1

19. Amongst the following, identify the species having an atom with +6 oxidation state.
(A) \(\mathrm{MnO}_{4}^{-}\)
(B) \(\mathrm{Cr}(\mathrm{OH})_{3}^{6-}\)
(C) \(\mathrm{NiF}_{6}^{2-}\)
(D) CrO₂Cl₂
Answer:
(D) CrO₂Cl₂

20. In which of the following compounds, the oxidation number of carbon is NOT zero?
(A) (CHCl)₂
(B) HCHO
(C) CH₃COOH
(D) CH₂Cl₂
Answer:
(C) CH₃COOH

21. The oxidation number of S in S₈, S₂F₂ and H₂S respectively are ………….
(A) 0, +1, -2
(B) +2, +1, -2
(C) 0, +1, +2
(D) +2, +1, -2
Answer:
(A) 0, +1, -2

22. The coefficients x, y, and z in the following balanced equation
xZn + \(\mathrm{yNO}_{3}^{-}\) + 10H+ → zZn²⁺ + \(\mathrm{NH}_{4}^{+}\) + 3H₂O are …………..
(A) 4, 1, 4
(B) 2, 2, 2
(C) 4, 2, 4
(D) 4, 4, 4
Answer:
(A) 4, 1, 4

23. For the redox reaction:
\(\mathrm{MnO}_{4}{ }^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)
The CORRECT coefficients of the reactants in the balanced reaction are ……………

Answer:
(A)

24. In the reaction
3CuO + 2NH₃ → N₂ + 3H₂O + 3Cu
the change of NH₃ to N₂ involve ……………..
(A) loss of 6 electrons per mol of N₂
(B) loss of 3 electrons per mol of N₂
(C) gain of 6 electrons per mole N₂
(D) gain of 3 electrons per mole N₂
Answer:
(A) loss of 6 electrons per mol of N₂

25. When KMnO₄ acts as an oxidising agent and ultimately forms \(\mathrm{MnO}_{4}^{2-}\), MnO₂, Mn₂O₃ and Mn²⁺, then the number of electrons transferred in each case is …………..
(A) 4, 3, 1, 5
(B) 1, 5, 3, 7
(C) 1, 3, 4, 5
(D) 3, 5, 7, 1
Answer:
(C) 1, 3, 4, 5

26. In electrochemical cell, the magnitude and direction of the electrode potential depends on which of the following?
(A) Nature of metal and ions
(B) Concentration of ions
(C) Temperature
(D) All of these
Answer:
(D) All of these

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