Question

25) Identify the element with the following electron configuration: [Kr5e14d5 A) Cr 8) Ni D) Tc E) Mo F) Re 26) How many unpaired electrons are present in the ground state sulphur atom? A) 0 B) 1 C) 2 D) 3 E) 4 F) 5 27) Identify the set of three quantum numbers that could represent the valence electrons in Sr A) -4,1 1, m= 1 B) n=4,1-0, m- 1 c)n 5,I= 1, m-1 -5,1-0, my = 0 D) E) n 5,1-1, m0 F)N=4,1-1, m0

Answer 1

25) Option E, Mo is the element having atomic number,Z= 42 which means electronic configuration is [Kr] 5s¹ 4d⁵

26) sulphur has atomic number, Z = 16

Electronic configuration : [Ne] 3s²3p⁴

P subshell is the valence subshell

as seen it has 2 unpaired electrons

บ้าน 1 11 ]

option C

27) Sr has atomic number 38

Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s²

4 quantum numbers

a) principle Quantum number, n : It is equal to number of shells. It can take any positive integral value except zero

It is written in front of every subshell or orbital.

b) azimuthal Quantum number, l : It is also equal to number of subshells. It can take value from 0 to n-1.

If l has value

Subshell a a jat of 20 - - - -

3) magnetic quantum number, m_{​​​​​​l} : It is equal to number of orbitals present in a subshell.

It can take value from +l to -l.

For example if l = 2, m_{​​​​​​l} can take the value of +2,+1,0, -1, -2.

4) spin Quantum number, m_{​​​​​​s}​​​: It can take only two values as an orbital can have only two electrons which can have opposite spins.

So value can be +1/2 or -1/2.

A orbital can be represented as

zou

where n: principle Quantum number

l : azimuthal Quantum number

x : number of electrons.

Now let's solve it for all valence orbital 5s² of Sr electronic configuration. Then match with the options given.

ne mems T5st 5 o o tylko

It matches with option D.