Which of the following orders are correct for the ionization energies?
(1) $B a < S r < C a$ (2) $S^{2 -} < S < S^{2 +}$ (3) $C < O < N$ (4) $M g < A l < S i$

1, 2 and 4

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1, 3 and 4

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1, 2 and 3

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1, 2, 3 and 4

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Solution

The correct option is C 1, 2 and 3
(1) Down the group size increases and therefore, ionization energies decrease. Hence the order is correct.
(2) Cation is smaller but anion is bigger than its parent atom. As charge on cation increases the ionic radius decreases. Similarly as charge on anion increases the ionic radius increases. $I E α \frac{1}{a t o m i c / i o n i c r a d i u s}$ .
Hence the order is correct.
(3) N has stable half filled configuration thus has higher ionization energy than O. Hence the order is correct.
(4) The correct order is $M g > A l > S i . M g (n s^{2})$ has higher ionization energy than $A l (n s^{2} n p^{1})$ because s-sub shell electrons have higher penetration power than that of p-sub shell electrons. Further across the period size decreases and nuclear charge increases and therefore, ionization energy increases.

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Similar questions

(1) B a < S r < C a

(2) S^{2 -} < S < S^{2 +}

(3) C < O < N

(4) M g < A l < S i

I

A l > S i > P > C l

I I

B e < M g < C a < S r

I I I

I < B r < C l < F

I V

{N a}^{+} < {M g}^{2 +} < {A l}^{3 +} < {S i}^{4 +}

A l^{3 +} < M g^{2 +} < N a^{+} < F^{-}

T e^{2 -} > I^{-} > C s^{+} > B a^{2 +}

F e^{3 +} < F e^{2 +} < F e^{4 +}

M g > A l > S i > P

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

N a^{+}

M g^{2 +}

A l^{3 +}

S i^{4 +}

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