Question

The formulae of copper sulphate is $CuS{{O}_{4}}$.What is the valency of copper if valency of $S{{O}_{4}}$is 2?

Answer 1

Hint: Cu is a symbol that is used for copper, belongs to d-block. The atomic number of Cu is 29. The $S{{O}_{4}}$ ion has a negative charge, i.e. $S{{O}_{4}}$ is in -2 state.

Complete step by step solution:
So, here we are provided with the data for the formulae of copper sulphate, which is $CuS{{O}_{4}}$ and we have to say that the Cu atom in, $CuS{{O}_{4}}$ is in which valence state.
As we know that Cu is a d –block element with atomic number 29.
The most characteristic feature of the d-block elements is that they show variable oxidation state, because the energy difference between ns orbital and (n-1) d orbital is very small the electron may be filled in both. And also electrons may be used from both to form compounds.
Since the Cu is having 29 electrons the expected electronic configuration was,
The electronic configuration of Cu=$\left[ Ar \right]3{{d}^{9}}4{{s}^{2}}$, but it exists as $\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}$
This electronic configuration is due to extra stability for the element due to completely filled d-orbitals.
But for doing the removal of electrons, to avoid confusion let’s take the electronic configuration of Cu as,
Cu=$\left[ Ar \right]3{{d}^{9}}4s$, as we all know that 4s orbital has the least energy and the removal of electrons will be done from the ns orbital first.
So in that case, the Cu is expected to have two valencies i.e. +1 and +2 valencies.
For exhibiting +1 electron, one electron is removed from the 4s orbital and configuration is,
\[C{{u}^{+1}}=\left[ Ar \right]3{{d}^{9}}4{{s}^{1}}\]
\[Cu\to C{{u}^{+1}}+1{{e}^{-}}\]
And by removing two electron it exists as,
$C{{u}^{2+}}=\left[ Ar \right]3{{d}^{9}}4{{s}^{0}}$
\[Cu\to C{{u}^{+2}}+2{{e}^{-}}\].
Now let’s calculate the valency of Cu in $CuS{{O}_{4}}$
Here as given in the question, $S{{O}_{4}}$ is in -2 state, so we assign -2 charge for $S{{O}_{4}}$ in the compound.
The oxidation number for a compound will be zero.
Give the oxidation state of Cu as x and solve the oxidation state for Cu,
$CuS{{O}_{4}}$ Has an oxidation state zero.
Hence, \[x+(-2)=0\]\[x+(-2)=0\]
X=2

Hence we got the valency of Cu as +2 in $CuS{{O}_{4}}$.

Note: For all neutral compounds the oxidation state is zero. All elements have an oxidation state zero. If in this compound the Cu was in +1 oxidation state then the formulae must have been like, $C{{u}_{2}}S{{O}_{4}}$ $C{{u}_{2}}S{{O}_{4}}$