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Chemistry NEET Questions — Free MCQ Practice

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Incorrect order of ionic size is :

A.

$L a^{3 +} > G d^{3 +} > E u^{3 +} > L u^{3 +}$

B.

$V^{2 +} > V^{3 +} > V^{4 +} > V^{5 +}$

C.

$T l^{+} > I n^{+} > S n^{2 +} > S b^{3 +}$

D.

$K^{+} > S c^{3 +} > V^{5 +} > M n^{7 +}$

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Explanation

(a) $L a^{3 +} > G d^{3 +} > E u^{3 +} > L u^{3 +}$ is incorrect due to lanthanide contraction because size will decrease from La+3 to Lu+3.

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Consider the following changes:

$M (s) \to M (g) . . . . . (i) M (s) \to M^{2 +} (g) + 2 e^{-} . . . . . (i i) M (g) \to M^{+} (g) + e^{-} . . . . . (i i i i) M^{+} (g) \to M^{2 +} (g) + e^{-} . . . . . . (i v) M (g) \to M^{2 +} (g) + 2 e^{-} . . . . . . (v)$

The second ionization energy of M could be calculated from the energy values associated with :

A.

1+3+4

B.

2-1+3

C.

1+5

D.

5-3

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Explanation

(d) Second ionization energy is amount of energy required to take out an electron from the monopositive cation.

Hence, $M_{(g)} \to M^{2 +} 2 e^{-} M_{(g)} \to M^{+} + e^{-}$

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Which of the following statements is/are wrong?

A.

van der Waals' radius of iodine is more than its covalent radius

B.

All isoelectronic ions belong to same period of the periodic table

C.

$I . E ._{1}$ of N is higher than that of O while $I . E ._{2}$ of O is higher than that of N

D.

The electron affinity N is almost zero while that of P is 74.3 kJ $m o l^{- 1}$

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Explanation

(b) In the isoelectronic species, all isoelectronic anions belong to the same period and cations to the next period.

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Consider the following conversions:

(i) $O_{(g)} + e^{-} \to O_{(g)}^{-} ∆ H_{1}$ (ii) $F_{(g)} + e^{-} \to F_{(g)}^{-}, ∆ H_{2}$

(iii) $C l_{(g)} + e^{-} \to C {l^{-}}_{(g)}, ∆ H_{3}$ (iv) $O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}, ∆ H_{4}$

That according to given information the incorrect statement is:

A.

$∆ H_{3}$ is more negative than $∆ H_{1} a n d ∆ H_{2}$

B.

$∆ H_{1} i s l e s s n e g a t i v e t h a n ∆ H_{2}$

C.

$∆ H_{1}, ∆ H_{2} a n d ∆ H_{3} a r e n e g a t i v e w h e r e a s ∆ H_{4} i s p o s i t i v e$

D.

$∆ H_{1} a n d ∆ H_{3} a r e n e g a t i v e w h e r e a s ∆ H_{2} a n d ∆ H_{4} a r e p o s i t i v e$

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Explanation

(d) Order of electron gain enthalpy : Cl>F>O

Second electron gain enthalpy for an element is always positive.

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Aqueous solutions of two compounds $M_{1} - O - H a n d M_{2} - O - H$ are prepared in two different beakers. If, the electronegativity of $M_{1} = 3.4, M_{2} = 1.2, O = 3.5 a n d H = 2.1,$ then the nature of two solutions will be respectively:

A.

acidic, basic

B.

acidic, acidic

C.

basic, acidic

D.

basic, basic

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Explanation

(a) The electronegativity difference between $M_{1}$ and O is 0.1, which indicates $M_{1} - O$ bond will be covalent, since $O — H$ bond having more ionic character thus bond will break and $H^{+}$ ions will release and acidic solution is formed. Whereas difference between electronegativity of $M_{2} — O$ bond is 2.3, thus, $M_{2} — O H$ bond will break. Hence, solution will be basic in nature.

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First three ionisation energies (in kJ/mol) of three representative elements are given below:

Element $I E_{1}$ $I E_{2}$ $I E_{3}$

P 495.8 4562 6910
Q 737.7 1451 7733
R 577.5 1817 2745

Then incorrect option is :

A.

Q: Alkaline earth metal

B.

P: Alkali metals

C.

R: s-block element

D.

They belong to same period

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Explanation

(c) R is p-block element, because difference between $I E_{2} a n d I E_{3}$ is not very high as compared to between $I E_{1} a n d I E_{2}; h e n c e s t a b l e o x i d a t i o n s t a t e o f R w i l l b e h i g h e r t h a n + 2 .$

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If the ionization enthalpy and electron gain enthalpy of an element are 275 and 86 kcal $m o l^{- 1}$ respectively, then the electronegativity of the element on the Pauling scale is:

A.

2.8

B.

0.0

C.

4.0

D.

2.6

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Explanation

(a) $I . E . + E . A . = 275 + 86 = 361 k c a l m o l^{- 1} = 361 \times 4.184 = 1510.42 k J m o l^{- 1} ∴ E l e c t r o n e g a t i v i t y = \frac{1510.42}{540} = 2.797 = 2.8$

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The incorrect statement is:

A.

The second ionisation energy of Se is greater than that of second ionisation energy of As

B.

The first ionisation energy of $C^{2 +}$ ion is greater than that of first ionisation energy of $N^{2 +}$ ion

C.

The third ionisation energy of F is greater than that of third ionisation energy of O

D.

Helogens have highest I.E. in respective period

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Explanation

(d)

$(a) \underset{4 p^{4}}{S e} \overset{I . E ._{1}}{\to} \underset{4 p^{3}}{S e^{+}} \overset{I . E ._{2}}{\to} \underset{4 p^{2}}{S e^{2 +}} \underset{4 p^{3}}{A s} \overset{I . E ._{1}}{\to} \underset{4 p^{2}}{A s^{+}} \overset{I . E ._{2}}{\to} \underset{4 p^{1}}{A s^{2 +}} (b) \underset{2 p^{2}}{C} \to \underset{2 p^{1}}{C^{+}} \to {\underset{2 s^{2}}{C}}^{2 +} \underset{2 p^{3}}{N} \to {\underset{2 p^{2}}{N}}^{+} \to {\underset{2 p^{1}}{N}}^{2 +} \underset{2 p^{4}}{O} \to {\underset{2 p^{3}}{O}}^{+} \to {\underset{2 p^{2}}{O}}^{2 +} (c) \underset{2 p^{5}}{F} \overset{I . E ._{1}}{\to} \underset{2 p^{4}}{F^{+}} \overset{I . E ._{2}}{\to} {\underset{2 p^{3}}{F}}^{2 +} \overset{I . E ._{3}}{\to} F^{3 +} \underset{2 p^{4}}{O} \overset{I . E ._{1}}{\to} \underset{2 p^{3}}{O^{+}} \overset{I . E ._{2}}{\to} {\underset{2 p^{2}}{O}}^{2 +} \overset{I . E ._{3}}{\to} O^{3 +} (d) I n r e s p e c t i v e p e r i o d, n o b l e g a s e s h a v e h i g h e r t h a n + 2 .$

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Consider the following ionisation reactions :

I.E. (kJ $m o l^{- 1}$ )

$A_{(g)} \to A_{(g)}^{+} + e^{-}, A_{1} B_{(g)}^{+} \to B_{(g)}^{2 +} + e^{-}, B_{2} C_{(g)}^{+} \to C_{(g)}^{2 +} + e^{-}, C_{2} B_{(g)} \to B_{(g)}^{+} + e^{-}, B_{1} C_{(g)} \to C_{(g)}^{+} + e^{-}, C_{1} C_{(g)}^{2 +} \to C_{(g)}^{3 +} + e^{-}, C_{3}$

If monovalent positive ion of A, divalent positive ion of B and trivalent positive ion of C have zero electron. Then incorrect order of corresponding I.E. is :

A.

$C_{3} > C_{2} > B_{2}$

B.

$B_{1} > A_{1} > C_{1}$

C.

$C_{3} > C_{2} > B_{2}$

D.

$B_{2} > C_{3} > A_{1}$

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Explanation

(d) $A \Rightarrow H (1 s^{1}) B \Rightarrow H e (1 s^{2}) C \Rightarrow L i (1 s^{2} 2 s^{1}) A_{1} = I E_{1} B_{2} = I E_{2} (B) B_{1} = I E_{1} (B) C_{2} = I E_{2} (C) C_{1} = I E_{1} (C) C_{3} = I E_{3} (C) B_{1} > A_{1} > C_{1} C_{3} > B_{2} > A_{1} C_{3} > C_{2} > B_{2} H e > H > L i L i^{2 +} H e^{+} H L i^{2 +} L i^{+} H e^{+} 1 s^{2} 1 s^{1} 2 s^{1} 1 s^{1} 1 s^{1} 1 s^{1} 1 s^{2} 1 s^{2} 1 s^{1}$

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Which of the following is the incorrect match for atom of element?

A.

$[A r] 3 d^{5} 4 s^{1} \to 4^{t h} p e r i o d, 6^{t h} g r o u p$

B.

$[K r] 4 d^{10} \to 5^{t h} p e r i o d, 12^{t h} g r o u p$

C.

$[R n] 6 d^{2} 7 s^{2} \to 7^{t h} p e r i o d, 3^{t h} g r o u p$

D.

$[X e] 4 f^{14} 5 d^{2} 6 s^{2} \to 6^{t h} p e r i o d, 4^{t h} g r o u p$

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Explanation

B

$(a) [A r] 3 d^{5} 4 s^{1} \to C r (24) \to 4^{t h} p e r i o d, 6^{t h} g r o u p (b) [K r] 4 d^{10} \to P d (46) \to 5^{t h} p e r i o d, 10^{t h} g r o u p (c) [R n] 6 d^{2} 7 s^{2} \to T h (90) \to 7^{t h} p e r i o d, 3 r d g r o u p (d) [X e] 4 f^{14} 5 d^{2} 6 s^{2} \to H f (72) \to 6^{t h} p e r i o d, 3^{r d} g r o u p$

Chemistry MCQs for NEET — Practice Questions with Answers

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