There is an element x ( atomic no. 19, atomic mass 40) (10) (a)what is its correct ground state orbital configuration (b) which group does it belongs to ? (c) which period does it belongs to (d) how many shells does it have ?

a)

b)  group no : 1

c) period no : 4

d) 4 shells

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies and thus the Electronic configuration for element X with atomic number of 19 and thus 19 electrons is:

b) The group for elements whose last electrons enter the s orbital is given by the number of electrons present in the valence s orbital. Thus the group number is 1.

c) The period of the element is given by the principal shell of the valence electron. Thus as last electron enters 4th main shell, the period number is 4.

d) The shells are given by the the value of principal quantum number and thus there are 4 values of principal quantum number i.e 1, 2 , 3 and 4 and thus there are 4 shells.

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.