Answer the question on basis of passage given below:

Nitrogen differs from the rest of the members of its group due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals. Nitrogen has unique ability to form $pπ-pπ$ multiple bonds with itself and with other elements having small size and high electronegativity (eg, C, O) Heavier elements of this group do not form $pπ-pπ$ bonds as their atomic orbitals are so large and diffused that they cannot have effective overlapping.

Nitrogen family show reducing characters. Least basic amongst the following is:

(A) $SbH_3$
(B) $BiH_3$
(C) $NH_3$
(D) $PH_3$

Choose the correct answer from the options given below:

(B) Only

The correct answer is Option (3) → (B) Only.

The basicity of hydrides in the nitrogen family (Group 15) decreases as you move down the group from nitrogen to bismuth. This trend can be explained by examining factors such as atomic size, electronegativity, and the availability of the lone pair of electrons.

Overview of Hydrides in Group 15

The hydrides of Group 15 elements include:

\(NH_3\) (Ammonia)

\(PH_3\) (Phosphine)

\(SbH_3\) (Stibine)

\(BiH_3\) (Bismuth Hydride)

Factors Influencing Basicity

Electronegativity: Basicity is largely determined by the ability of the atom to donate its lone pair of electrons. Higher electronegativity means a stronger attraction for electrons, making the lone pair less available for bonding.

Atomic Size: As the size of the central atom increases down the group, the bond lengths also increase. This results in a weaker bond with hydrogen, making it less effective at donating its lone pair.

Lone Pair Availability: The availability of the lone pair for bonding decreases as you go down the group, as larger atoms have their electrons in higher energy levels, making them less accessible.

Detailed Analysis of Each Hydride

\(NH_3\) (Ammonia)

Structure: Trigonal pyramidal geometry with a lone pair on nitrogen.

Electronegativity: Nitrogen is highly electronegative (3.04 on the Pauling scale), which enhances its ability to donate the lone pair.

Basicity: Strongly basic due to effective donation of the lone pair, making it capable of forming bonds with protons (\( \text{H}^+ \)). It can also engage in hydrogen bonding, which stabilizes its structure.

2. \(PH_3\) (Phosphine)

Structure: Pyramidal geometry with a lone pair on phosphorus.

Electronegativity: Phosphorus is less electronegative than nitrogen (2.19), reducing the availability of its lone pair for donation.

Basicity: Less basic than ammonia. While it can still act as a weak base, it does not form hydrogen bonds effectively, leading to decreased stability.

3. \(SbH_3\) (Stibine)

Structure: Similar pyramidal geometry with a lone pair on antimony.

Electronegativity: Antimony has an even lower electronegativity (2.05), further reducing the availability of its lone pair.

Basicity: Even less basic than phosphine. The larger size and lower electronegativity contribute to the poor donation ability of the lone pair.

4. \(BiH_3\) (Bismuth Hydride)

Structure: Pyramidal with a lone pair on bismuth.

Electronegativity: Bismuth is the least electronegative (1.88) among these elements, meaning the lone pair is very poorly available for donation.

Basicity: The least basic of all, as the larger atomic radius and low electronegativity make the lone pair very difficult to donate.

Summary of Basicity Order
The basicity of these hydrides can be summarized in the following order:

Most Basic: \(NH_3\) > \(PH_3\) > \(SbH_3\) > \(BiH_3\)

This trend is evident in their ability to react with acids and form ammonium or similar ions.

Conclusion
Given this analysis,  \(BiH_3\) (Bismuth Hydride) is the least basic hydride among the options provided, due to its larger size, lower electronegativity, and poor availability of the lone pair for donation. Therefore, the correct answer is: (B) Only \((BiH_3)\)