Question

The percentage of $$\pi$$-character in the orbitals forming $$P - P$$  bonds in $${P_4}$$ is

A. 25
B. 33
C. 50
D. 75  
Answer :   75
Solution :
In $${P_4}$$ ,the $$P - P$$   linkage is formed by $$s{p^3} - s{p^3}$$  hybridised orbital overlapping. So the percentage of $$\pi $$ -character will be $$75\% $$

Releted MCQ Question on
Inorganic Chemistry >> P - Block Elements

Releted Question 1

The reddish brown coloured gas formed when nitric oxide is oxidised by air is

A. $${N_2}{O_5}$$
B. $${N_2}{O_4}$$
C. $$N{O_2}$$
D. $${N_2}{O_3}$$
Releted Question 2

The temporary hardness of water due to calcium carbonate can be removed by adding —

A. $$CaC{O_3}$$
B. $$Ca{\left( {OH} \right)_2}\,$$
C. $$CaC{l_2}$$
D. $$\,HCl$$
Releted Question 3

Which of the following is most stable to heat

A. $$HCl$$
B. $$HOCl$$
C. $$HBr$$
D. $$HI$$
Releted Question 4

White $$P$$ reacts with caustic soda. The products are $$P{H_3}$$ and $$Na{H_2}P{O_2}.$$   This reaction is an example of

A. Oxidation
B. Reduction
C. oxidation and reduction
D. Neutralisation

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P - Block Elements


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