# PERIODIC TRENDS IN RADIUS OF ATOM AND ION IIT JEE - NEET

## Atomic radii of fluorine and neon in Angstrom units are respectively given by:

(IIT JEE 1987)

a) 0.72, 1.60

b) 1.60, 1.60

c) 0.72, 0.72

d) None of these

Logic and Solution:

The reported radii of noble gas elements are "van der Waals radii", which are 40% more than the actual atomic radii. Hence neon's atomic radius must be much more than that of fluorine.

Note: It is not possible to get covalent and metallic radii for noble gases since they do not form bonds.

Conclusion:

Above fact is reflected in option "a".

## Related questions

### 2) Which of the following is the smallest in size?

(IIT JEE 1988)

1) N3-

2) O2-

3) F-

4) Na+

Logic & solution:

 Atom N O F Na Atomic number, Z (or) # protons 7 8 9 11 # electrons 7 8 9 11 Add   3 electrons  | | | ∇ Add   2 electrons  | | | ∇ Add   1 electron | | | ∇ Subtract  1 electron  | | | ∇ Ion N3- O2- F- Na+ # electrons 10 10 10 10 Z /# electrons 7/10 = 0.7 8/10 = 0.8 9/10 = 0.8 11/10 = 1.1

The ratio between no. of protons (Z) to number of electrons correlates effective nuclear attraction. Greater this value, greater is the attraction and smaller is the size.

For example, in Na+ ion, there is 1.1 proton for each electron. Hence the attraction is maximum and the ion is smallest among the given species.

However, in N3- ion, there is only 0.7 proton for each electron. Hence the attraction is minimum and the ion is largest.

Generalization: In iso-electronic species, the atomic size decreases with increase in the atomic number.

Conclusion:

The smallest is Na+

### 3) The correct order of radii is:

(IIT JEE 2000)

a) N < Be < B

b) F- < O2- < N3-

c) Na < Li < K

d) Fe3+ < Fe2+ < Fe4+

Logic & solution:

Option a: Incorrect. Boron is smaller than Beryllium. The atomic size decreases across a period from left hand side to right hand side. The correct order is: N < B < Be.

 group 2 group 13 group 15 Be B N -----------> Atomic radius decreases.

Option b: Correct. The ionic radii increases with decrease in the effective nuclear charge. Already explained.

 N3- O2- F- 10 10 10 7/10 = 0.7 8/10 = 0.8 9/10 = 0.8 --------------------------> Size decreases with increase in proton number

Option c: Incorrect. The atomic size increases from top to bottom in a group. The correct order is: Li < Na < K.

 Li | | | Atomic size increases down the group | | ∇ Na K

Option d: For ions of same element, the size should decrease with increase in the positive charge. The correct order must be: Fe3+ > Fe2+ > Fe4+.

Conclusion:

The correct option is "b".

### 4) The correct order of ionic radii is:

(IIT JEE 1999)

a) Ti4+ < Mn7+

b) 37Cl- < 35Cl-

c) K+ > Cl-

d) P3+ > P5+

Logic & solution:

Option a: Incorrect. Ti4+ and Mn7+ are isoelectronic. Both of them contain 18 electrons in it. The size decrease with increase in the positive charge these species.

 Ion Ti4+ Mn7+ # protons(Z) 22 25 # electrons 18 18 Z/#electrons 22/18=1.22 25/18=1.38

Option b: Incorrect. These ions belong to two different isotopes of same element. Since the proton number is same in isotopes, the nuclear attraction is also same. Hence their radii must be equal.

Option c: Incorrect. K+ and Cl- are isoelectronic species but the number of protons are more in K+. Hence it is smaller in size than Cl-.

 Ion K+ Cl- # protons(Z) 19 17 # electrons 18 18 Z/#electrons 19/18=1.05 17/18=0.94

Option d: Correct. For different ions of same element, size decreases with increase in the positive charge.

Conclusion:

The correct option is "d"

### 5) Which of the following is correct order of ionic radii?

(EAMCET 2000-M)

1) Na+ < Mg2+ < Al3+ < Si4+

2) Al3+ < Si4+ < Na+> Mg2+

3) Si4+ < Al3+ > Mg2+ > Na+

4) Na+ > Mg2+ > Al3+ > Si4+

Logic & solution:

 Atom Na Mg Al Si Atomic number, Z (or) # protons 11 12 13 14 # electrons 11 12 13 14 Subtract  1 electron  | | | ∇ Subtract  2 electron  | | | ∇ Subtract  3 electron  | | | ∇ Subtract  4 electron | | | ∇ Ion Na+ Mg2+ Al3+ Si4+ # electrons 10 10 10 10 Z /# electrons 11/10 = 1.1 12/10 = 1.2 13/10 = 1.3 14/10 = 1.4

For isoelectronic ions, the radius decreases with increase in positive charge.

Conclusion:

The correct order of ionic radii is  Na+ > Mg2+ > Al3+ > Si4+ as given in option "4".

### 6) The radius of La3+ (atomic number : La=57) is 1.06 Å. Which one of the following given values will be closest to the radius of Lu3+ (atomic number : Lu =71) ?

(AIEEE 2003)

a) 1.60 Å

b) 1.40 Å

c) 1.06 Å

d) 0.85 Å

Logic & solution:

Lanthanum, La and Lutetium, Lu belong to lanthanoids (4f block elements of inner transition elements). Lanthanum is the first element and Lutetium is the last one. In lanthanoids, the atomic and ionic radii decrease with increase in atomic number. It is called as lanthanoid contraction.

Conclusion:

Since 0.85 Å is the only option that is less than 1.06 Å, the correct option is 'd'.

## Homework

1) What is Lanthanoid contraction?

2) Why zirconium and Hafnium have nearly same atomic sizes?