Question

1. This one has multiple parts

a) How many valence electrons does an atom of the element whose atomic number is 36 have? Enter only a number for your answer.
b) What is the period number for the element described in the previous question? Enter only a number for your answer.
c) What is the chemical symbol for the element described in the previous questions?
d) What is the name for the element described in the previous question?
e) How many total electrons does an atom of Rubidium have? Enter only a number for your answer.

2. Answer the following questions about the central atom in the optimized Lewis structure for carbon tetraiodide (CI₄)

a) Number of covalent bonds attached to the central atom =  
b) Number of lone pairs of electrons on the central atom =  
c) Formal charge on the central atom (format: 0,+1,-1, etc.) =  
d) The electron groups have a  ---Select--- linear trigonal planar tetrahedral trigonal bipyramidal octahedral geometry
e) The hybridization of the electron groups is  ---Select--- sp sp2 sp3 sp3d sp3d2
f) The molecule or ion has a  ---Select--- linear trigonal planar bent tetrahedral trigonal pyramidal trigonal bipyramidal see-saw T-shaped octahedral square pyramidal square planar shape

Answer 1

a) It has 8 valence electrons. Because it will be in group 18 due to it's atomic number and group 18 elments have 8 valence electrons

b) It's in 4th period.

c) Kr is the chemical symbol.

d) name of element is Krypton.

e) Rubidium has 37 electrons. Because it's atomic number is 36.

Ques 2)

a) 4 covalent bonds are attached to central atom. This can be understood from it's structure, the Carbon atom is bound to 4 Iodine atoms so 4 covalent bonds are present.

b) Central atoms doesn't have any lone pair. This is because the electronic configuration is 2s2 2p2 ​​​​​​. On excitation, it is 2s1 2p3 all these 4 unpaired electron are involved in bonding with Iodine atom. So it doesn't have any lone pair of electrons.

c) Formal charge = No. Of valence electron - (non-bonded electron + No. of bonds)

For Carbon in CI4,

Formal charge = 4 - (0+ 4) = 0

d) Electron geometry is 3D arrangement of atom around central atom. For CI4, the geometry is tetrahedral because it's hybridisation is sp​​​​​3

e) Hybridisation is sp​​​​​3​​​​. This can be calculated by steric Number method.

Steric number = No of sigma bonds+ No of non- conjugated Lone pair.

Steric number= 4 + 0. =4 (As no lone pair)

For steric nunber 4, hybridisation is sp​​​​​3​​​​sp3

f) since the molecule doesn't contain any lone pair so structure is not distorted and the shape will be tetrahedral due to sp​​​​​3 geometry.