# The correct statement among the following is : Option 1)$\left ( SiH_{3} \right )_{3}N$ is pyramidal and less basic than $\left (CH_{3} \right )_{3}N$.  Option 2)$\left ( SiH_{3} \right )_{3}N$nis planar and less basic than  $\left (CH_{3} \right )_{3}N$.Option 3)$\left ( SiH_{3} \right )_{3}N$ is planar and more basic than $\left (CH_{3} \right )_{3}N$.  Option 4)$\left ( SiH_{3} \right )_{3}N$ is pyramidal and more basic than $\left (CH_{3} \right )_{3}N$.

Electronegativity of Nitrogen family -

Gradually decreases down the group. More electronegative than group 14

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Positive oxidation state of Nitrogen family -

Shows +3 and +5 oxidation states (except N2)

Stability of +3O.S increases down the group due to inert pair effect

- wherein

+3 when only p electron are involved and +5 when all valence five (s and p) electrons are involved

Negative oxidation state of Nitrogen family -

Exhibits -3 O.S, decreases on moving down the group

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$\left ( SiH_{3} \right )_{3}N$ is planar & less basic than $\left ( CH_{3} \right )_{3}N$.

Because, lone pair of Nitrogen-atom is in unhybridised p-orbital & it overlaps with d-orbital of si-atom, thus planar in nature.

$\left ( SiH_{3} \right )_{3}N$ is less basic than $\left ( CH_{3} \right )_{3}N$ as lone pair of electrons on nitrogen-atom is in conjugation with all Si-atoms.

Option 1)

$\left ( SiH_{3} \right )_{3}N$ is pyramidal and less basic than $\left (CH_{3} \right )_{3}N$.

Option 2)

$\left ( SiH_{3} \right )_{3}N$nis planar and less basic than  $\left (CH_{3} \right )_{3}N$.

Option 3)

$\left ( SiH_{3} \right )_{3}N$ is planar and more basic than $\left (CH_{3} \right )_{3}N$.

Option 4)

$\left ( SiH_{3} \right )_{3}N$ is pyramidal and more basic than $\left (CH_{3} \right )_{3}N$.

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