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Hint: To find the number of electrons in $B{{r}^{-}}$, you should first know that how many electrons are present in bromine and then, add one electron in it and then, those resultant number of electrons will be present in $B{{r}^{-}}$. Now you can easily answer the statement.
Complete Solution :
- Bromine is a gas which comes under the category of halogens and belongs to the p-block elements of the periodic table.
- It lies in the 17th group and 4th period of the periodic table.
- It has an atomic number as 35 and the mass number as 80. It has the electronic configuration as: $3{{d}^{10}}4{{s}^{2}}4{{p}^{5}}$.
- Bromine has seven electrons in its outermost valence shell and has a strong tendency to gain one electron to acquire the fully filled electronic configuration of the nearest noble gas i.e. krypton.
Now considering the statement as:
- The number of electrons present in an atom is always equal to the atomic number of that very element.
- So, therefore, the number of electrons present in bromine = $35$.
Since, $B{{r}^{-}}$ carries negative charge i.e. it means it has gained one electron to complete its octet.
Thus, the total number of electrons present in $B{{r}^{-}}$ = $35 + 1 = 36$.
Note: Bromine being a halogen has seven electrons in its outermost shell and requires only one electron to complete its octet. So, bromine has a strong tendency to gain one electron from the other electron donating species and thus, it acts as a strong Lewis acid.
Complete Solution :
- Bromine is a gas which comes under the category of halogens and belongs to the p-block elements of the periodic table.
- It lies in the 17th group and 4th period of the periodic table.
- It has an atomic number as 35 and the mass number as 80. It has the electronic configuration as: $3{{d}^{10}}4{{s}^{2}}4{{p}^{5}}$.
- Bromine has seven electrons in its outermost valence shell and has a strong tendency to gain one electron to acquire the fully filled electronic configuration of the nearest noble gas i.e. krypton.
Now considering the statement as:
- The number of electrons present in an atom is always equal to the atomic number of that very element.
- So, therefore, the number of electrons present in bromine = $35$.
Since, $B{{r}^{-}}$ carries negative charge i.e. it means it has gained one electron to complete its octet.
Thus, the total number of electrons present in $B{{r}^{-}}$ = $35 + 1 = 36$.
Note: Bromine being a halogen has seven electrons in its outermost shell and requires only one electron to complete its octet. So, bromine has a strong tendency to gain one electron from the other electron donating species and thus, it acts as a strong Lewis acid.
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