Which of the following order is wrong ?

N H_{3} < P H_{3} < A s H_{3} - a c i d i c

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L i < B < B e < C - 1 s t I P

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A l_{2} O_{3} < M g O < N a_{2} O < K_{2} O - b a s i c

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L i^{+} < N a^{+} < K^{+} < C s^{+} - i o n i c r a d i u s

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Solution

The correct option is B $L i < B < B e < C - 1 s t I P$
$L i, B e, B$ and $C$ are present across a period. In a period from left to right, ionisation potential increases.

But in case of Be and B, Be has higher ionization potential than B due to stable configuration of Be.

So, the correct order of IP of given elements is:
$L i < B < B e < C$

Suggest Corrections

Similar questions

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

The correct order of reactivity of alkalimetals towards $H_{2}$ is

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

S i F_{4} < S i C l_{4} < S i B r_{4} < S i I_{4}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

N H_{3} > S b H_{3} > A s H_{3} > P H_{3}

M g O, K_{2} O, {N a}_{2} O, C a O

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