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John Bhargava
Asked: 2022-11-08T08:41:02+05:30 In:

Nitrogen forms `N_(2)` but phosphorous when forms `P_(2)` gets readily converted into `P_(4)` because
A. triple bond is present between phosphorus atoms
B. `p pi -d pi ` bonding is weak
C. `p pi- p pi` bonding is strong
D. multiple bond is formed easily.

Nitrogen forms `N_(2)` but phosphorous when forms `P_(2)` gets readily converted into `P_(4)` because
A. triple bond is present between phosphorus atoms
B. `p pi -d pi ` bonding is weak
C. `p pi- p pi` bonding is strong
D. multiple bond is formed easily.
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  1. Correct Answer – B
    `P_(2)` is unstable because the `p pi -d pi ` bonding in it is weak. It is due to larger atomic size of P and N. thus `P_(2)` is converted into `P_(4)` in which there is no `p pi -dpi` bonding.

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Chinmay Kashyap
Asked: 2022-11-06T20:04:31+05:30 In:

Nitrogen forms `N_(2)` but phosphorous when forms `P_(2)` gets readily converted into `P_(4)` because
A. `p pi – p pi` bonding is strong in phosphorus
B. `p pi – p pi` bonding is weak in phosphorus
C. triple bond is present in phosphorus
D. single P – P bond is weaker than N -N bond.

Nitrogen forms `N_(2)` but phosphorous when forms `P_(2)` gets readily converted into `P_(4)` because
A. `p pi – p pi` bonding is strong in phosphorus
B. `p pi – p pi` bonding is weak in phosphorus
C. triple bond is present in phosphorus
D. single P – P bond is weaker than N -N bond.
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  1. Correct Answer – B
    `p pi – p pi` bonding in nitrogen is strong hence it can form triple bond with another N. Single N – N bond is weaker than P – P bond due to high interionic repulsion of non-bonding electrons. Hence, `N equiv N` is stable and P 2 is not.

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