Question

How many valence electrons are in $ IC{l_5}? $

Answer 1

Hint: The valence shell is the last shell or the outermost shell of every element. We know that every element has a unique electronic configuration which describes the distribution of electrons in various shells or orbits or energy levels of every atom. The last of the electronic configuration denotes the valence shell and the electrons in the outermost shell is called the valence electrons.

Complete Step by Step answer
We know that every element or atom in the periodic table has a unique atomic number and an electronic configuration based on which they are arranged in the periodic table. The electronic configuration specifies a lot about the electronic arrangement of the atom and its orbitals. Out of which the final terms of the electronic configuration usually specify about the outermost shell of the element and the electrons in that shell. The outermost shell is called a valence shell and the electrons in this shell are called the valence electrons.
The valence electrons have a great influence on the element and its nature to share electrons with other elements and the formation of compounds. We also know that the given compound $ IC{l_5} $ consists of $ I $ iodine and chlorine $ \left( {Cl} \right) $ both from the halogen group of the periodic table. We know that the common electronic configuration of halogens are: $ n{s^2}n{p^5} $ .Thus the number of valence electrons are seven.
Now in the given compound we have 1 iodine atom and 5 other chlorine atoms, thus each atom has 7 valence electrons. Thus, the total valence electrons for the 6 atoms in the compound will be, $ 6 \times 7 = 42{\text{electrons}} $ .

Note
In the given compound we know that halogen atoms are monovalent and hence the central iodine atom forms 5 single bonds with each chlorine atom thus leaving all the chlorine atoms with 3 lone pairs of electrons and the central iodine atom with 1 lone pair of electrons. Thus, forming a square pyramidal structure.