Question

Increasing order of bond length in NO, \[N{O^ + }\]and \[N{O^ - }\] is:
A.\[NO > N{O^ - } > N{O^ + }\]
B.\[N{O^ + } < NO < N{O^ - }\]
C.\[NO < N{O^ + } < N{O^ - }\]
D.\[NO < N{O^ + } = N{O^ - }\]

Answer 1

Hint:We know that the bond length is a measure of the distance between the nuclei of two chemically bonded atoms in a molecule and it is approximately equal to the sum of the covalent radii of the two bonded atoms. For the covalent bonds, the bond length is inversely proportional to the bond order which means that higher bond orders will result in the stronger bonds, which are accompanied by stronger forces of attraction that holds the atoms together. Short bonds are a consequence of these strong forces of attraction.

Complete answer:
The bond strength increases going from \[N{O^ - }\] to \[N{O^ + }\], and the bond length consequently shortens going from \[N{O^ - }\] to \[N{O^ + }\].
\[N{O^ - } > NO > N{O^ + }^{}\]
 \[N{O^ + }\] has \[0\;\pi *\] antibonding electrons.
NO has \[1\;\pi *\] antibonding electron.
\[N{O^ - }\] has \[2\;\pi *\] antibonding electrons.
As the number of antibonding electrons increases, the N−O bond weakens, having acquired an antibonding character (which as the name suggests, goes against making a bond).
Thus, the bond strength increases going from \[N{O^ - }\] to \[N{O^ + }\], and the bond length consequently shortens going from \[N{O^ - }\] to \[N{O^ + }\].
Hence, increasing order of bond length in NO, \[N{O^ + }\] and \[N{O^ - }\] is \[N{O^ + } < NO < N{O^ - }\]

Therefore, the correct answer is option (B).
Note:
The periodic trends that can be observed in the bond lengths of elements are similar to the periodic trends in the atomic radii of the elements as the bond length decreases across the period, and it increases down the group. Bonded atoms vibrate due to thermal energy available in the surroundings and they are typically in the range of 100-200 pm (1-2 Å).