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Bond Angle

Which of the ...

Question

Which of the following orders of bond angle is correct?

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

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H_{2} O < H_{2} S < H_{2} S e < H_{2} T e

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O F_{2} < H_{2} O < O C l_{2}

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

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Solution

The correct option is A $N H_{3} < P H_{3} < A s H_{3} < S b H_{3}$
value bond theory;
$O F_{2} < H_{2} O < O C l_{2}$
This is the correct order of bond angle.
Because according to the theory electronegativity of the surrounding atom increases,bond angle decreases.

Suggest Corrections

Similar questions

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}

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}

H_{2} O < H_{2} T e < H_{2} S e < H_{2} S

H_{2} O > H_{2} T e > H_{2} S e > H_{2} S

H_{2} O > H_{2} S > H_{2} S e > H_{2} T e

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

H_{2} E

(E = O, S, S e, T e a n d P o

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