Silver atoms have completely filled d orbitals $\left( {4{d^{10}}} \right)$ in its ground state. How can you say that it is a transition element?

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Hint: We perceive that silver is a transition component and the chemical symbol of silver is \[Ag\] and the atomic number of silver is\[47\]. It is a delicate, white, glossy progress metal; it shows the most elevated electrical conductivity, warm conductivity, and reflectivity of any metal.

Complete step by step answer:
The transition components are those components having a half filled d or f subshell in any basic oxidation state. The expression "progress components" most ordinarily alludes to the d-block change components. $Ag$ has a completely packed \[4d\] orbital \[\left( {4{d^{10}}5{s^1}} \right)\] in its ground state. Presently, silver presentations two oxidation states \[( + 1)\] and $( + 2)$ . In the \[ + 1\] oxidation express, an electron is taken out from the s-orbital. Notwithstanding, in the \[ + 2\] oxidation express, an electron is eliminated from the d-orbital. Accordingly, the d-orbital now gets deficient \[\left( {4{d^9}} \right)\] then, silver is a transition component.

Note:
Now, let us discuss the variable valency.
The capacity of the progress components to clarify variable valency is commonly credited to the accessibility of more electrons in the orbital $\left( {n - 1} \right)d$ which is nearer to the utmost $ns$ orbital in energy levels. Variable valency is commonly found experiencing significant transition components of the periodic table and it occurs as the subsequent last shell of progress metal and it isn't loaded up with electrons.