Answer
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Hint: Octet rule- Octet rule states that the elements which are having \[1 - 7\] number of electrons in its outermost shell take part in the phenomenon of the bond formation and they also acquire the electronic configuration of the nearest noble gas.
Complete step by step solution:
As we know that the number of electrons in the outermost shell of the nitrogen are five $(5)$, so they will need three $(3)$ more electrons to complete the requirement of the eight $(8)$ electrons in its outermost shell.
In option (A) the given compound is ammonia and in these three electrons of the nitrogen is shared by the three atoms of hydrogens, so now the nitrogen is having eight electrons in its last valence shell and hydrogen is also having two electrons in its last shell. Hence here the octet rule is followed.
In option (B) given compound is methane in which carbon is having four electrons in its outermost shell and they are shared by the other four electrons of the hydrogen atoms, thus carbon is having eight electrons and hydrogen is having two electrons in the outermost shell. Hence octet rule is followed.
Similarly, ${\text{C}}{{\text{O}}_{\text{2}}}$ is also following the octet rule.
But for ${\text{NO}}$there is nitrogen${\text{N}}$ atom, in which there are five electrons in the valence shell of the element its two electrons are shared with the two electrons of the oxygen to form two covalent bonds. Hence total electrons that take part to form the bond are seven$(7)$, but as the octet rule electrons must be $8$, so here we see that the octet rule is violated here for the ${\text{NO}}$ molecule.
Hence in all the four options only option (D) is not following the octet rule.
So, option (D) is the correct answer.
Note: This comes in the limitations of the octet rule that there is some incomplete octet of the central atom in some compounds, that means the number of the electrons surrounding the central atom is less than eight. This case is basically with the elements having less than four electrons in its valence shell. Some more examples of this limitation are ${\text{LiCl,Be}}{{\text{H}}_{\text{2}}}{\text{,BC}}{{\text{l}}_{\text{3}}}$.
Complete step by step solution:
As we know that the number of electrons in the outermost shell of the nitrogen are five $(5)$, so they will need three $(3)$ more electrons to complete the requirement of the eight $(8)$ electrons in its outermost shell.
In option (A) the given compound is ammonia and in these three electrons of the nitrogen is shared by the three atoms of hydrogens, so now the nitrogen is having eight electrons in its last valence shell and hydrogen is also having two electrons in its last shell. Hence here the octet rule is followed.
In option (B) given compound is methane in which carbon is having four electrons in its outermost shell and they are shared by the other four electrons of the hydrogen atoms, thus carbon is having eight electrons and hydrogen is having two electrons in the outermost shell. Hence octet rule is followed.
Similarly, ${\text{C}}{{\text{O}}_{\text{2}}}$ is also following the octet rule.
But for ${\text{NO}}$there is nitrogen${\text{N}}$ atom, in which there are five electrons in the valence shell of the element its two electrons are shared with the two electrons of the oxygen to form two covalent bonds. Hence total electrons that take part to form the bond are seven$(7)$, but as the octet rule electrons must be $8$, so here we see that the octet rule is violated here for the ${\text{NO}}$ molecule.
Hence in all the four options only option (D) is not following the octet rule.
So, option (D) is the correct answer.
Note: This comes in the limitations of the octet rule that there is some incomplete octet of the central atom in some compounds, that means the number of the electrons surrounding the central atom is less than eight. This case is basically with the elements having less than four electrons in its valence shell. Some more examples of this limitation are ${\text{LiCl,Be}}{{\text{H}}_{\text{2}}}{\text{,BC}}{{\text{l}}_{\text{3}}}$.
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