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The ion that is isoelectronic with \[CO\] is:
A. \[{O_2}^ - \]
B. \[{N_2}^ + \]
C. \[C{N^ - }\]
D. \[{O_2}^ + \]

Answer
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Hint:An ion is an atom or a molecule which has a net electric charge. The charge on the atom may be positive or negative.

Complete step by step answer: The term ion refers to an electronically charged species other than the neutral one. This species is either positively or negatively charged. The positive charged species is termed as cation and the negative charged species is termed anion.
The cation is formed by loss of electrons and the anion is formed by gain of electrons. The cation is labeled as \[{X^ + }\] and the anion is labeled as \[{Y^ - }\] ion.
The term isoelectronic is related to the number of electrons present on the respective atom or ion. Isoelectronic atoms, ions, or molecules are the ones which possess identical electronic structure and the number of valence electrons is equal. The prefix iso means the same and electronic means electrons. Isoelectronic species exhibit similar chemical properties.
The number of electrons for each atoms or ions is calculated as follows:
For \[CO\] : number of electrons of \[C\]+ number of electrons of \[O\] = $6 + 8 = 14electrons$
For \[{O_2}^ - \]: \[2{\text{ }} \times \] number of electrons of \[O\] + \[1\] = $2 \times 8 + 1 = 17electrons$
For \[{N_2}^ + \]: \[2{\text{ }} \times \] number of electrons of \[N\] - \[1\] = $2 \times 7 - 1 = 13elecrtons.$
For \[C{N^ - }\]: number of electrons of \[C\]+ number of electrons of \[N\] + \[1\] = $6 + 7 + 1 = 14electrons.$
For \[{O_2}^ + \]: \[2{\text{ }} \times \] number of electrons of \[O\] – \[1\] = $2 \times 8 - 1 = 15electrons.$
Of the given four ions the \[C{N^ - }\]ion has \[14\] electrons which are the same with the \[CO\].
Thus option C is the correct answer.

Note:
Other than atoms, molecules or ions the isoelectronic is also applied in case of amino acids. Some compounds like \[C{H_3}COC{H_3}\] and \[C{H_3}N = NC{H_3}\] have the same number of electrons but they are not isoelectronic.