Question

An element with electronic arrangement as $2,8,18,1$ will exhibit which of the following stable oxidation states?

• A. $+2$ and $+4$
• B. $+1$ and $+2$
• C. $+2$ to $+7$
• D. $+1$ only

Answer 1

The correct option is D $+1$ only

Electronic configuration $2,8,18,1$ is $1s^{2}2s^{2}2p^{6}3s^3{p}^{6}3d^{10}4s^1$

Therefore by lossing 1 electron it can attain stable configuration will completely filled orbitals. Only +1 oxidation state will be exhibited.