How do you evaluate $-{{7}^{2}}$?

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. Finally, we find the value for indices number 2 and express it in its simplest form.

Complete step-by-step solution:
We have been given to find the value of $-{{7}^{2}}$.
We first try to explain the concept of square of a real number.
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.
Therefore, ${{a}^{n}}=\underbrace{a\times a\times a\times ....\times a\times a}_{n-times}$.
The value of $n$ can be any number belonging to the domain of real numbers.
Similarly, the value of $a$ can be any number belonging to the domain of real numbers.
In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
Now for our given indices we reframe it as $-{{7}^{2}}=-\left( {{7}^{2}} \right)$.
We complete the multiplication inside the brackets to get ${{7}^{2}}=7\times 7=49$.
Therefore, $-{{7}^{2}}=-\left( {{7}^{2}} \right)=-49$.
The solution for $-{{7}^{2}}$ is $-49$.

Note: We need to remember the difference between $-{{7}^{2}}$ and ${{\left( -7 \right)}^{2}}$.
We previously got $-{{7}^{2}}=-49$.
But in case of ${{\left( -7 \right)}^{2}}$, we have ${{\left( -7 \right)}^{2}}=\left( -7 \right)\times \left( -7 \right)=49$.
The absence of the bracket changes the sign of the solution.