The atomic numbers of three elements A, B and | vedclass

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(C) The electronic configuration of the elements is as follows:Element $A$ (atomic number $9$): $2, 7$. It needs $1$ electron to complete its octet,so

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$C$

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(C) The electronic configuration of the elements is as follows:Element $A$ (atomic number $9$): $2, 7$. It needs $1$ electron to complete its octet,so it forms an anion.Element $B$ (atomic number $10$): $2, 8$. It has a complete octet and is a noble gas,so it is inert.Element $C$ (atomic number $13$): $2, 8, 3$. It has $3$ valence electrons. Metals tend to lose their valence electrons to achieve a stable octet configuration,thereby forming a cation ($+$ve ion).Therefore,element $C$ will form a cation by losing $3$ electrons: $C (2, 8, 3) \rightarrow C^{3+} (2, 8) + 3e^-$.

The atomic numbers of three elements $A, B$ and $C$ are $9, 10$ and $13$ respectively. Which of them will form a cation?

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