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Which of the following compound is most acidic?

$C l_{2} O_{7}$

$P_{4} O_{10}$

$S O_{3}$

$B_{2} O_{3}$

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Explanation

(a) Oxidation state of non- metal increases acidic nature of oxide increase $\overset{+ 7}{C l_{2} O_{7}}$ is most acidic.

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A compound contains three elements A, B and C, if the oxidation number of A =+2, B =+5 and C =-2, the possible formula of the compound is :

$A_{3} {(B_{4} C)}_{2}$

$A_{3} {(B C_{4})}_{2}$

$A_{2} {(B C_{3})}_{2}$

$A B C_{2}$

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Explanation

(b) Sum of oxidation numbers of all atoms in a neutral molecule is zero.

$\to A_{3} {(B_{4} C)}_{2} : + 2 \times 3 + 2 \times [4 \times 5 - 2] = 42 \neq 0 \to A_{3} {(B C_{4})}_{2} : + 2 \times 3 + 2 \times [+ 5 - 8] = 0 \to A_{2} {(B C_{3})}_{2} : + 2 \times 2 + 2 \times [+ 5 - 6] = + 2 \neq 0 \to A B C_{2} : + 2 + 5 + (- 2) \times 2 = + 3 \neq 0$

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If period number and group number of any representative element(s) are same then which of the following statement is incorrect regarding such type element(s) in their ground state? (Period number and group number are according to modern form of periodic table)

The possible value of principal quantum number is 2

The possible value of azimuthal quantum number is zero

The possible value of magnetic quantum number is 1

The species could be paramagnetic

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Explanation

(i) H 1 1 $1 s^{1}$

(ii) Be 2 2 $1 s^{1} 2 s^{2}$

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How does the energy gap between successive energy levels in an atom vary from low to high n values?

All energy gaps are the same

The energy gap decreases as n increases

The energy gap increases as n increases

The enenrgy gap changes unpredictably as n increases

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Explanation

(b) As value of n increases, energy gap decreases due to increasing $Z_{e f f .}$ on valence shell.

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Which of the following properties of the alkaline earth metals increase from Be to Ba?

(i) Atomic radius (ii) Ionisation energy (iii) Nuclear charge

(i) and (ii)

(i) and (iii)

(ii) and (iii)

(i), (ii) and (iii)

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Explanation

(b) Atomic radius and nuclear charge increases from top to bottom because number of shell and atomic mass increases down the group.

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Incorrect order of ionic size is :

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

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

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

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

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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 :

1+3+4

2-1+3

1+5

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?

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

All isoelectronic ions belong to same period of the periodic table

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

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:

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

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

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

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

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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:

acidic, basic

acidic, acidic

basic, acidic

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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