Question

An element having atomic mass 37 contains 20 neutrons. Write the electronic configuration of the element and locate the position of this element in periodic table.

Answer 1

Answer:

The electronic configuration of the atom is

1s2 2s2 2p6 3s2 3s 2 3p6 4s1 3d10.

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Answer 2

Given :

• Atomic mass = 37
• Number of neutrons = 20

To find :

• Electronic configuration of the element
• Position in periodic table

Solution :

• The atomic mass of an element is the mass of its nucleus .
• Since the nucleus contains protons and neutrons of considerable mass , the atomic mass of an element is the sum of mass of number of protons and the mass of number of neutrons .
• Each proton and neutron has mass = 1u , so the atomic mass of an element is the sum of number of protons and neutrons.

Atomic mass = Neutrons + Protons

37 = 20 + Protons

Protons = 37 - 20

Protons = 17

The number of protons = Number of electrons = Atomic number .

Therefore , the electronic configuration of element 17 is :

• 2 , 8 , 7 ( in K L M N electronic configuration)
• 1s² 2s² 2p⁶ 3s² 3p⁵ (in spdf electronic configuration)

The position of the element is in period 3 because it has last electron in third shell .

Group 17 because the last shell contains 7 electrons and group number is equal to 10 + total number of electrons in last shell ( if n > 4) .

Therefore ,

• Period : 3
• Group : 17
• Element : Chlorine ( Cl)