Explore topic-wise InterviewSolutions in Current Affairs.

Linear, 180⁰

Trigonal planar and bond angle 120⁰

In the formation of ethene molecule, one of the sp² hybrid orbitals of carbon atom overlaps axially with sp² hybridised orbital of another carbon atom to form C–C sigma bond. While the other two sp² hybrid orbitals of each carbon atom are used for making sp² –s sigma bond with two hydrogen atoms. The unhybridised orbital (2px or 2py) of one carbon atom overlaps sidewise with the similar orbital of the other carbon atom to form weak π bond.

sp³ tetrahedral

Total no. of students = 50,000

Number of students like Pulses and Chapatti = 60% of 50,000

= 50,000 x 60/100 = 30,000

Number of students like Vegetable and Chapatti = 25% of 50,000

= 50,000 x 25/100

= 12,500

Remaining students

= 50,000 – (30,000 + 12,500)

= 50,000 – 42,500

= 7,500

∴ No. of students like Khichdi = 7,500

∴ Percentage of students like Khichdi

= 7500/50000 x 100

= 15%

Hence, 15% students like Khichdi.

(b) four times

The correct option is (b) 5.

1. discount

2. VAT

3. Compound Interest

4. x₂/y₂

5. decreased.

1. Utility

2. Physical transfer

3. Contractual Income

4. Wealth

5. Science of wealth

6. Lionel Robbins

7. Alfred Marshall

8. the ‘Arthashastra’

9. “Queen of Social Science”

10. Social Science/Abstract Science

11. Chanakya or Vishnugupta

12. Sir Ragnar Frisch

Wrong Pair:

Adam Smith – Scarcity Oriented Definition

Correct pair

(c) 1 – c; 2 – b; 3 – d; 4 – a

Wrong Pair: 

Prof. Alfred Marshall – Father of Economics

Steps involved in balancing a Chemical Equation:

Step 1: Write the correct skeleton equation.

Step 2: Start with the compound that has the maximum atoms or maximum kinds of atoms and the atoms present in it are balanced first.

Step 3: Balance elements that appear only once on each side of the arrow first. Then balance elements that appear more than once on a side.

Step 4: Elementary substances are balanced last of all.

Step 5: If required the whole equation is multiplied by some suitable number in order to make all the coefficients whole numbers.

Correct Answer is: (a) Advertising

(d) Kaveripumpattinam

Kautilya Who advocated maintenance of perfect balance between State management and people’s welfare through trading activities.

Correct Answer is: (a) Kautilya

Boats like ‘Padgu’ , ‘Thimil’ , ‘Thom’ , ‘Ambu’ ‘Odampunai’ , etc., were used to cross rivers for domestic trade.

Boats like ‘Kalam’ , ‘Marakalam’ , ‘Vangam’ , ‘Navai’ , etc., were used for crossing oceans for foreign trade.

Trade in Sangam period was both internal and external.

Day market was called as Nalangadi.

Try our new mock tests series with multiple attempts and calculate your performance unless you get fully satisfied with your preparation level. The d-and f-block elements are an important section of chemistry in class 12. Start practicing The d-and f-Block Elements Class 12 Mock Test and score maximum in the actual board exam of class 12. Give mock tests daily and evaluate your performance and understand that where you want to give more efforts.

Mock test for class 12 Chemistry are prepared as per the newest syllabus and Exam Pattern. Students can solve these Class 12 Mock test with Answers and assess their preparation level.

Students are advised to practice the Class 12 online Mock test of The d-and f-Block Elements to reconsider the essential ideas and find out about the kind of questions that may appear on the board exam. Students can practice these The d-and f-Block Elements Mock test with Answers and evaluate their level of preparation.

Click here for regular practice: - Class 12 The d-and f-Block Elements Mock Test

Students can solve these The d-and f-Block Elements MCQ Questions with Answers and evaluate their level of preparation. Students are advised to practice the Class 12 Chemistry MCQ Questions of The d-and f-Block Elements with Answers to reconsider the essential ideas and find out about the kind of questions that may appear on the board exam.

MCQ Questions for class 12 Chemistry with Answers are prepared as per the newest syllabus and Exam Pattern. Students can solve these Class 12 MCQ Questions with Answers and assess their preparation level.

Practice MCQ Question for Class 12 Chemistry chapter-wise

1. Shape of d-orbital is

(a) spherical
(b) dumb bell
(c) double dumb bell
(d) none of these

2. Electronic configuration of alkaline earth element is

(a) ns²
(b) ns¹
(c) np⁶
(d) ns⁰

3. Lanthanoid contraction is due to increase in

(a) atomic number
(b) effective nuclear charge
(c) atomic radius
(d) valence electrons

4. Which one of the following is diamagnetic ion?

(a) Co²⁺
(b) Ni²⁺
(c) Cu²⁺
(d) Zn²⁺

5. Colour of transition metal ions are due to absorption of some wavelength. This results in

(a) d-s transition
(b) s-s transition
(c) s-t/transition
(d) d-d transition

6. The melting points of Cu, Ag and Au follow the order

(a) Cu > Ag > Au
(b) Cu >Au > Ag
(c) Au > Ag > Cu
(d) Ag > Au > Cu

7. In which of the following pairs of ions, the lower oxidation state in aqueous solution is more stable than the other?

(a) Ti⁺, Ti³⁺
(b) Cu⁺, Cu³⁺
(c) Cr²+, Cr³⁺
(d) V²+, VO²⁺ (V⁴)

8. Zr and Hf have almost equal atomic and ionic radii because of

(a) diagonal relationship
(b) lanthanoid contraction
(c) actinoid contraction
(d) belonging to the same group

9. Malachite is an ore of 

(a) Silver
(b) Mercury
(c) Magnesium
(d) Copper

10. The chief ore of Hg is

(a) Pyrolusite
(b) Bauxite
(c) Galena
(d) Cinnabar

11. The transition element which shows the highest oxidation state is: 

(a) Iron
(b) Vanadium
(c) Manganese
(d) Chromium

12. Gun metal is

(a) Cu + Zn
(b) Cu + Sn + Zn
(c) Cu + Sn
(d) Zn + Sn

13. German silver is an alloy of:

(a) Fe, Cr, Ni
(b) Ag, Cu Au
(c) Cu, Zn, Ni
(d) Cu, Zn, Sn

14. The composition of duralumin is

(a) Al 94%, Mg 6%
(b) Cu 56%, Zn 24%, Ni 20%
(c) Cu 95%, Al 5%
(d) Al 95%, Cu 4%, Mn 0.5%, Mg 0.5%

15. Which of the following is not an ore of iron?

(a) limonite
(b) Cassiterite
(c) magnetite
(d) none of these

16. Lanthanide for which +II and +III oxidation states are common is

(a) La
(b) Pr
(c) Ce
(d) Eu

17. F₂ is formed by reacting K₂MnF₆ with

(a) SbF₅
(b) MnF₃
(c) KSbF₆
(d) MnF₄

18. Which of the following radioisotopes is used as anticancerous?

(a) Na -24
(b) C -14
(c) U -235
(d) Co -60

19. The colour of copper sulphide is

(a) Blue
(b) Black
(c) Red
(d) Green

20. What is wrong about transition metals?

(a) Diamagnetic
(b) Paramagnetic
(c) Form complexes
(d) Shows variable oxidation state

21. The last electron in d-block elements goes to

(a) (n-1) d
(b) nd
(c) np
(d) (n-1) s

22. The elements which exhibit both vertical and horizontal similarities are

(a) inert gas elements
(b) representative elements
(c) rare elements
(d) transition elements

23. An atom has electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s² in which group would it be placed?

(a) Fifth
(b) Fifteenth
(c) Second
(d) Third

24. Number of unpaired electrons in Ni²⁺ (Z = 28) is

(a) 4
(b) 2
(c) 6
(d) 8

25. Transition metals usually exhibit highest oxidation states in their

(a) chlorides
(b) fluorides
(c) bromides
(d) iodides

Answer :

1. Answer : (c) double dumb bell

Explanation: The ml values for five d orbitals are -2, -1, 0, +1, and +2 i.e., we can say d-subshell has five orientations. All these d-orbitals have the same energy and are called degenerate orbitals. Hence, we can say d-orbitals have double dumbbell-shaped.

2. Answer : (a) ns²

Explanation: The general electronic configuration of alkaline earth elements is ns². These elements have two valence electrons in the outermost s shell.

3. Answer : (b) effective nuclear charge

Explanation: Lanthanoid contraction effect results from poor shielding of nuclear charge (nuclear attractive force on electrons) by 4f electrons so effective nuclear charge increases; the 6s electrons are drawn towards the nucleus, thus resulting in a smaller atomic radius.

4. Answer : (d) Zn²⁺

Explanation: Zn²⁺ = [Ne 3s² 3p⁶ 4s⁰ 3d¹⁰]  It is diamagnetic, so no unpaired e⁻s.

5. Answer : (d) d-d transition

Explanation: Colour of transition metal ions are due to absorption of the same wavelength. This results in d−d transition.

6. Answer : (b) Cu >Au > Ag

Explanation: Hence the melting point decreases from copper to silver, but the melting point of gold becomes even higher than silver due to its higher density that happened due to contraction.

7. Answer : (a) Ti⁺, Ti³⁺

Explanation: Ti⁺ ions are more stable than Ti³⁺ and thus, Ti³⁺ ions change to Ti⁺ ions thereby acting as oxidising agents.

8. Answer : (b) lanthanoid contraction

Explanation: Zr and Hf have almost similar atomic and ionic radii it can be explained on the basis of shielding effect . Due to lanthanide contraction. Moreover, the electrons which are present in inner shells, shields outer electrons from nuclear charge, after that they experience a low effective nuclear charge.

9. Answer : (d) Copper

Explanation: Malachite, a minor ore but a widespread mineral of copper, basic copper carbonate, Cu₂CO₃(OH)₂. Because of its distinctive bright green colour and its presence in the weathered zone of nearly all copper deposits, malachite serves as a prospecting guide for that metal.

10. Answer : (d) Cinnabar

Explanation: Cinnabar is an ore of mercury.

11. Answer : (c) Manganese

Explanation: Manganese has the highest oxidation state because the number of unpaired electrons in the outermost shell is more, i.e. 3d⁵ 4s².

12. Answer : (b) Cu + Sn + Zn

Explanation: Gunmetal, also called G Metal, variety of bronze, formerly used for ordnance. Modern admiralty gunmetal is composed of 88 percent copper, 10 percent tin, and 2 percent zinc and is used for gears and bearings that are to be subjected to heavy loads and low speeds.

13. Answer : (c) Cu, Zn, Ni

Explanation: German Silver is an alloy of copper, zinc and nickel, sometimes also containing lead and tin.

14. Answer : (d) Al 95%, Cu 4%, Mn 0.5%, Mg 0.5%

Explanation: Duralumin is an alloy 95% aluminium, 4% copper, 0.5% manganese and 0.5% magnesium.

15. Answer : (b) Cassiterite

Explanation: Cassiterite was the main tin ore throughout ancient history and remains the most important source of tin today. Thus we can say that cassiterite isn't an ore of iron.

16. Answer : (d) Eu

Explanation: Lanthanides show variable oxidation states. The most stable oxidation state of Lanthanides is +3. They also show +2 and +4 oxidation states due to the presence of either half filled or completely filled or empty 4f sub shell. Due to stable half filled configuration Eu shows +2 configuration.

17. Answer : (a) SbF₅

Explanation: F₂ is the formed by reacting K₂MnF₆ with Sb₅. Reaction is as follows:

K₂MnF₆ + 2SbF₅ → 2KSbF₆ + MnF₃ + \(\frac{1}{2}\)F₂

18. Answer : (d) Co -60

Explanation: Co-60 is used as anti-cancerous among the given radioactive isotopes. It emits \(\beta\)− particles and energetic gamma rays, for which reason it is used in radiation therapy. 

19. Answer : (b) Black

Explanation: When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper sulphide is obtained and the sulphuric acid so formed remains in the solution.

20. Answer : (a) Diamagnetic

Explanation: Transition elements have one or more unpaired electrons. Due to the presence of these unpaired electrons, they show paramagnetism. Thus, transition metals are not diamagnetic. They are good conductors of heat and electricity. They form complexes and show variable oxidation state as d orbitals are gradually filled up in the transition series.

21. Answer : (a) (n-1) d

Explanation: In these elements, the last electron enters the d orbital of the penultimate shell i.e. the last electron goes to (n-1) d orbital. Hence these elements are named as d-block elements.

22. Answer : (d) transition elements

Explanation: Elements which exhibit both vertical and horizontal similarities are transition elements. They show vertical similarity because of same electronic configuration. They show horizontal similarity because of similar size.

23. Answer : (a) Fifth

Explanation: The outermost shell and one shell prior to it in d subshell, has total 5 electron so it will be placed in fifth group in transition element or 'd' block of periodic table.

24. Answer : (b) 2

Explanation: Nickel has two unpaired electrons in the 3d sublevel. When nickel ionizes to the +2 state it loses the two electrons in the 4s, so there are still two unpaired electrons in Ni²⁺.

25. Answer : (b) fluorides

Explanation: The highest oxidation states of transition metals are shown by fluorides and oxides as fluorine and oxygen are the most electronegative elements. The highest oxidation state shown by any transition metal is +8.

Click here to practice  MCQ Question for The d-and f-Block Elements Class 12 

These MCQ Questions Online Tests cover the whole of Control and Coordination. Our experts are prepared MCQ Questions for Class 10 based on the latest exam pattern. Students can refer to these Class 10 Science MCQ Questions of Control and Coordination with Answer for their exam preparation. 

MCQ Questions of Control and Coordination are taken from all important topics of the chapter and answer just as an explanation is given underneath each question.

Practice Class 10 Science MCQ Questions of Control and Coordination

1. Which of the following acts as both endocrine and exocrine glands?

(a) Adrenal
(b) Pituitary
(c) Ovaries
(d) Pancreas

2. Which hormone regulates the ionic balance in the body?

(a) Glucagon
(b) Thyroxine
(c) Testosterone
(d) Vasopressin

3. Impulse flows from ___.

(a) dendrite to axon in same neuron
(b) axon to dendrite
(c) nodes of ranvier to axon
(d) none

4. ____ are multipolar and conduct impulse out from spinal cord.

(a) neuroglia
(b) neurolemma
(c) schwann cells
(d) motor neuron

5. Which of the following statements is correct about receptors?

(a) Gustatory receptors detect taste while olfactory receptors detect smell
(b) Both gustatory and olfactory receptors detect smell
(c) Auditory receptors detect smell and olfactory receptors detect taste
(d) Olfactory receptors detect taste and gustatory receptors smell

6. In a neuron, conversion of electrical signal to a chemical signal occurs at/in

(a) cell body
(b) axonal end
(c) dendritic end
(d) axon

7. Posture and balance of the body is controlled by

(a) cerebrum
(b) cerebellum
(c) medulla
(d) pons

8. Spinal cord originates from

(a) cerebrum
(b) medulla
(c) pons
(d) cerebellum

9. The gap between two neurons is called a:

(a) dendrite
(b) synapse
(c) axon
(d) impulse

10. The brain is responsible for:

(a) thinking
(b) regulating the heart beat
(c) balancing the body
(d) all of the above

11. Cartisons are secreted by:

(a) pancreatic gland
(b) adrenal gland
(c) thyroid gland
(d) pituitary gland

12. Female sex hormone is:

(a) estrogen
(b) androgen
(c) insulin
(d) adrenalin

13. The main function of the plant hormone called abscisic acid is to:

(a) Increase the length of cells
(b) Promote cell division
(c) Inhibit growth
(d) Promote growth of stem and roots

14. The growth of tendrils in pea plants is due to the:

(a) Effect of sunlight on the tendril cells facing the sun
(b) Effect of gravity on the part of tendril hanging down towards the earth
(c) Rapid cell division and elongation in tendril cells that are away from the support
(d) Rapid cell division and elongation in tendril cells in contact with the support

15. The plant hormone which triggers the fall of mature leaves and fruits from the plant body is:

(a) Auxin
(b) Gibberellin
(c) Abscisic acid
(d) Cytokinin

16. The stimulus in the process of thigmotropism is:

(a) Touch
(b) Gravity
(c) Light
(d) Chemical

17. The number of pairs of nerves which arises from the spinal cord is:

(a) 21
(b) 31
(c) 41
(d) 51

18. Reflex actions are mediated through

(a) brain
(b) effectors
(c) spinal cord
(d) receptors

19. The highest coordinating centre in the human body is

(a) spinal cord
(b) heart
(c) brain
(d) kidney

20. Iodine is necessary for the synthesis of which of the following hormone?

(a) Adrenaline
(b) Auxin
(c) Thyroxine
(d) Insulin

Answer key

1. (d) Pancreas
2. (d) Vasopressin
3. (a) dendrite to axon in same neuron
4. (d) motor neuron
5. (a) Gustatory receptors detect taste while olfactory receptors detect smell
6. (b) axonal end
7. (b) cerebellum
8. (b) medulla
9. (b) synapse
10. (d) all of the above
11. (b) adrenal gland
12. (a) estrogen
13. (c) Inhibit growth
14. (c) Rapid cell division and elongation in tendril cells that are away from the support
15. (c) Abscisic acid
16. (a) Touch
17. (b) 31
18. (c) spinal cord
19. (c) brain
20. (c) Thyroxine 

Click here to practice more MCQ Questions from Chapter Control and Coordination Class 10 Science

Eustachian valve is present at the opening of inferior vena cava while Thebeisan valve is present near the opening of coronary sinus.

The process of formation of RBCs is called erythropoiesis.

The decrease in the number of white blood cells is called leucopenia whereas decrease in the number of red blood cells is called erythrocytopenia.

The hematocrit is ratio of the volume of RBCs to total blood volume of blood.

Using angle sum property for ΔPRT, we obtain

∠PRT + ∠RPT + ∠PTR = 180º

40º + 95º + ∠PTR = 180º

∠PTR = 180º − 135º

∠PTR = 45º

∠STQ = ∠PTR = 45º (Vertically opposite angles)

∠STQ = 45º

By using angle sum property for ΔSTQ, we obtain 

∠STQ+ ∠SQT + ∠QST = 180º 

45º + ∠SQT + 75º = 180º 

∠SQT = 180º − 120º

∠SQT = 60º

Order is 3 × 3 because it is product of two matrices having order 3 × 1 and 1 × 3.

\(a_{23} = \frac{1}{3}|3\times2+3|\)

\(= \frac{1}{3}|-6+3|\)

\(=\frac{1}{3}\times3\) = 1

If A is identity matrix, 

Then,  A = I₂

\(\begin{bmatrix} cos\,\alpha & -sin\,\alpha \\[0.3em] sin\,\alpha & cos\,\alpha \\[0.3em] \end{bmatrix} \)= \(\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix} \)

On equating, we get

⇒ cos α = 1,sin α = 0

⇒ α = 0

Using equality of two matrices,

7y = – 21

⇒ y = – 3

2x – 3y = 11

⇒ 2x + 9 = 11

⇒ x = 1

∴ x = 1, y = – 3

∵ a_ij = \(\frac{i}{j}\) 

⇒ a₁₂ = \(\frac{1}{2}\)  [ Here i = 1 and j = 2]

It is given that

f(x) = 1/(1 + x²)

By differentiating w.r.t. x

f’(x) = –2x/(1 + x²)² ≤ 0

Therefore, f(x) is increasing function for all x ≤ 0.

Consider f(x) = 3^x

The domain of f(x) is R.

f’(x) = 3^x ln(3)

3^x is always greater than 0 and ln(3) is also + ve.

Overall f’(x) is >0 means strictly increasing in its domain i.e. R.

f(x) = (x - 1/x)

domain of function is R - {0}

f'(x) = (1+1/x²)

f’(x) ∀ x ∈ R is greater than 0.

Consider f(x) = x³ - 15x²+75x - 50

Domain of the function is R.

f’(x) = 3x² - 30x+75

= 3(x² - 10+25)

= 3(x - 5)(x - 5)

= 3(x - 5)²

f’(x) = 0 for x = 5

for x<5

f’(x)>5

and

for x>5

f’(x)>5

we can see throughout R the derivative is +ve but at x = 5 it is 0 so it is increasing.

It is given that

f(x) = [x³ + 1/x³]

By differentiating w.r.t. x

f’(x) = 3x² – 3x^-4

By taking 3 as common

f’(x) = 3[x² – 1/x⁴]

Taking LCM

f’(x) = 3[(x⁶ – 1)/x⁴] = 3[(x²)³ – 1]/x⁴

On further calculation

f’(x) = [3(x² – 1) (x⁴ + x² + 1)]/x⁴

So we get

f’(x) = [3(x – 1) (x + 1) (x⁴ + x² + 1)]/x⁴ < 0 for x ∈ (- 1, 1).

Therefore, f(x) is decreasing function on ]-1, 1[.

It is given that

f(x) = [3/x + 5]

By differentiating w.r.t. x

f’(x) = 3(-1/x²) = -3/x² < 0 for all value of R

Therefore, f(x) is decreasing for all x ∈ R, where x ≠ 0.

f(x) = -x³+3x² - 3x+4.

f’(x) = -3x²+6x - 3

f’(x) = -3(x² - 2x+1)

f’(x) = -3(x -1)²

As f’(x) has –ve sign before 3

⇒ f’(x) is decreasing over R.

It is given that

f(x) = x/sin x

By differentiating w.r.t. x

f’(x) = (sin x. 1 – x. cos x)/(sin x)² = (sin x – x cos x)/(sin x)²

We know that

f’(x) > 0

By substituting the values

(sin x – x cos x)/(sin x)² > 0

So we get

(sin x – x cos x) > 0

Here tan x > x which is true on ] 0, π/2[.

Let r be the radius of the base and h the height of a cylinder.

The surface area is given by,

S = 2 π r² + 2 π rh

h = S - 2πr²/2πr ………(1)

Let V be the volume of the cylinder.

Therefore, V = πr²h

V = πr² \((\frac{S-2\pi r^3}{2\pi r})\) …….Using equation 1

V = \(\frac{Sr-2\pi r^3}{2}\)

Differentiating both sides w.r.t r, we get,

\(\frac{dV}{dr}=\frac{S}{2}\) - 3πr² ……….(2)

For maximum or minimum, we have,

 \(\frac{dV}{dr}\) = 0

⇒ \(\frac{S}{2}\) - 3πr² = 0

⇒ S = 6πr²

2πr² + 2πrh = 6πr²

h = 2r

Differentiating equation 2, with respect to r to check for maxima and minima, we get,

 \(\frac{d^2V}{dr^2}\) = - 6πr < 0

Hence, V is maximum when h = 2r or h = diameter

Consider f(x) = \(\frac{1}{(1+x^2)}\),

f'(x) = \(-\frac{2x}{(1+x^2)^2}\)

for x ≥ 0,

f’(x) is -ve.

hence function is decreasing for x ≤ 0

Correct option is: A) axioms

Correct option is: D) Statement

(a) It is given that f(x) = x²

By differentiating w.r.t x

f’(x) = 2x > 0 for x ∈ (0, ∞)

Here f’(x) > 0 which means 2x > 0

So we get x > 0

(b) We know that

If f’(x) < 0 we get x < 0

(c) It is given that f(x) = x² is strictly increasing on [0, ∞[ and strictly decreasing on ]- ∞, 0[

So it is neither strictly increasing nor strictly decreasing on the whole real line.

It is given that

f(x) = 5x – 2

By differentiating w.r.t. x

f’(x) = 5 > 0 for all x > 0

Therefore, f(x) is strictly increasing function for all x > 0.