Covalent radius:

Covalent radius: It is defined as one half of the distance between the nuclei (inter nuclear distance) of two covalently bonded like atoms in a homo nuclear diatomic molecule is called the covalent radius of that.

(A) For Homo atomic Molecules: The covalent radius (r_A) of atom A in a molecule A₂ may be given as:

The distance between nuclei of two single covalently bonded atoms in homo diatomic molecules is equal to the sum of covalent radii of both the atoms.

Illustrative example (1): A given compound A₂ whose total d_A-A is 1.4 A⁰. The atomic (covalent) radius of an atom is.

Solution:  We known that

(B) For Heteroatomic Molecules: In a hetero diatomic molecules AB where the electro negativity of atoms A and B are different, the experimental values of inter nuclear distance d_A-B is less than the theoretical value (r_A +r_B).

(1) Stevenson & Schoemaker Equation (1941):

Covalent radius of heterogeneous molecule like A-B etc determine by Stevenson & Schromaker Equation, if atoms are formed different type of covalent bond i.e. on atom is more electronegative than the other combined atom. Then the covalent radius is calculated by the relation given by Stevenson & Schoemaker, given as:

For a diatomic Hetero molecule:

Bond Length (l_A-B) = r_A + r_B- 0.09(X_A-X_B)

Where X_A= Electronegativity of more electronegative atom

Where X_B= Electronegativity of less electronegative atom

Illustrative examples 

(1) The electronegativity of F and H are 4.0 and 2.1 respectively. The percentage ionic character in H and F bond is.

(2) A given compound A2 whose total dA-A is 1.4 A0. The atomic (covalent) radius of an atom is.

(3) A given compound AB whose electronegative difference is 1.9 . Atomic radius of A and B are 4 and 2 Angstroms the distance between A and B means dA-B is ?

(4)If covalent radius of A is 0.99 A0 and bond length is 1.46 A0 Calculate the covalent radius of B. XA and XB are 3.5 and 2.5 respectively.

(5) The C–C single bond length is 1.54 Angstroms and that of Cl–Cl is 1.98 Angstroms. If the electronegativity of Cl and C are 3.0 and 2.5 respectively, the C–Cl bond-length will be equal to ?

(2) Puling equation: If the electro negativities of the two atoms A and B are X_A and X_B   respectively then

                    Bond Length (d_A-B) =( r_A +r_B)-(C₁X_A-C₂X_B)

Where C₁ and C₂ are the Stevenson’s coefficients for atoms A and B respectively 

Related Questions:

(1) What are the Amphoteric metals ? gives Examples.

(2) Name of total metalloids present in periodic table ?

(3) Total numbers of elements which are liquid at normal temperature is ?

(4) What is Mendeleev's periodic table ? give important features and draw back of Mendeleev's table.

(5) What is atomic density ? give the periodicity of atomic density in periods and groups.

(6) What is atomic volume ? and what is periodicity of atomic volume in groups and periods ?

(7) Why there are 2, 8 and 8 elements in first, second and third periodic of periods table respectively ? Explain.

(8) In alkali metal group which is the strongest reducing agent in aqueous solution and why?

(9) The electron affinity of sulphur is greater than oxygen. Why?

(10) The first ionization energy of carbon atom is greater than that of boron atom, whereas reverse is true for the second ionization energy. Explain.

(11) The electronegativities of B, Al, Ga are 2.0, 1.5, 1.6 respectively. The trend is not regular. Explain.

(12) Li2CO3 decomposes on heating but other alkali metal carbonates don’t. Explain.

(13) Of all noble metals, gold has got a relatively high electron affinity. Explain.

(14) What are the increasing order of ioni radii of first group elements in water ?

(15) What are the increasing order of molar conductivity of first group elements in water ?

A given compound A2 whose total dA-A is 1.4 A0. The atomic (covalent) radius of an atom is.

 Solution:  We known that

Solved Questions:

(1) The electronegativity of F and H are 4.0 and 2.1 respectively. The percentage ionic character in H and F bond is.

(2) A given compound AB whose electronegative difference is 1.9 . Atomic radius of A and B are 4 and 2 Angstroms the distance between A and B means dA-B is ?

(3)If covalent radius of A is 0.99 A0 and bond length is 1.46 A0 Calculate the covalent radius of B. XA and XB are 3.5 and 2.5 respectively.

(4) The C–C single bond length is 1.54 Angstroms and that of Cl–Cl is 1.98 Angstroms. If the electronegativity of Cl and C are 3.0 and 2.5 respectively, the C–Cl bond-length will be equal to ?

(5) Stevenson & Schromaker Equation: Determination covalent radius of Heterogeneous Molecules.

(6) How to calculate percentage (%) ionic character in covalent compounds?

If covalent radius of A is 0.99 A0 and bond length is 1.46 A0 Calculate the covalent radius of B. XA and XB are 3.5 and 2.5 respectively.

 (A) 0.35 A⁰              (B) 0.56 A⁰              (C) 1.98A⁰               (D) 10.48A⁰

The covalent radius of B is calculated by the relation given by Stevenson & Schoemaker, given as:

For a diatomic Hetero molecule:

Bond Length (l_A-B) = r_A + r_B- 0.09(X_A-X_B)

Where X_A= Electronegativity of more electronegative atom

Where X_B= Electronegativity of less electronegative atom

Bond Length=1.46 A⁰ = 0.99 A⁰ + r_B - 0.09(3.5-2.5)

 Hence r_B = 0.56 A⁰ 

Solved Questions:

(1) The electronegativity of F and H are 4.0 and 2.1 respectively. The percentage ionic character in H and F bond is.

(2) A given compound AB whose electronegative difference is 1.9 . Atomic radius of A and B are 4 and 2 Angstroms the distance between A and B means dA-B is ?

(3) The C–C single bond length is 1.54 Angstroms and that of Cl–Cl is 1.98 Angstroms. If the electronegativity of Cl and C are 3.0 and 2.5 respectively, the C–Cl bond-length will be equal to ?

(4) Stevenson & Schromaker Equation: Determination covalent radius of Heterogeneous Molecules.

(5) How to calculate percentage (%) ionic character in covalent compounds?

Why is AlCl3 in covalent and AlF3 in ionic?

Since Fluoride (F-) is smaller in size, it polarisability is less and therefore it is having more ionic character. Whereas chloride (Cl-) being larger in size is having more polarisability and hence more covalent character.

Related Questions:

(1) What are the hydrolysis products of urea ?

(2) CCl4 can not be hydrolysed but SiCl4 can be. Why?

(3) Why PCl3 hydrolysed while NCl3 can not be hydrolysed?

(4) Why SF6 behave inert towards hydrolysis?

(5) Arrange in increasing order of extent of hydrolysis [ CCI4, MgCI2, AICI3, PCl5, SiCI4].

(6) Although Sulphur contain vacant d-orbital but SF6 does not under go hydrolysis. Why ?

(7) Arrange the silicon halides into decreasing order of Lewis acids Character? SiF4, SiCl4, SiBr4, SiI4

Why Back bonding does not take place in P(SiH3)3 instead of phosphorus has lone pair and Silicon has vacant d-orbital but in N(SiH3)3 molecule back bonding does take place?

Conditions for back bonding:

(1) Both of the atoms bonded with back Bonding are must be present in 2nd-2nd or 2nd-3rd period. 4rth period onward back bonding does not take place.

(2) One of the atoms has lone pair (donor atom) and another (acceptor atom) have vacant Orbital and direction of back Bonding depends upon vacant Orbital.

(3) The donor atom must have localized donatable electron pair and there should be inter electronic repulsion (smaller size). In general these are later half second period P - block elements (F, O, N and C).

(4) The acceptor atom must have low energy empty orbital which generally are np or nd orbitals. Small and similar size orbital’s favour overlap.

(5) Back bonding is a weak pi bond thus only effective overlapping will be form back bonding.

(6) Back bonding is found to be effective and considerable in following type of overlapping.

 (i) 2p-2p

(ii) 2p-3p

(iii) 2p-3d 

(7) The extent of overlapping order is [2p-2p> 2p-3d >2p-3p]

(8) dx²-y² and dx² Orbital’s does not participate in back Bonding. 

Solution:

In case of  trisilyl phosphine [P(SiH₃)₃]  Phosphorous (P) being larger in atomic size so it does not face much interelectronic repulsion and so, not much eager to donate it’s loan pair electron and comfortable with it. So, it will not donate to Si atom in this case. Thus, back bonding will not take place in trisilyl phosphine [P(SiCH₃)₃].

But in the case of trisilyl amine [N(SiH₃)₃], the nitrogen (N) atom is smaller in atomic size leading to high interelectronic repulsion, so it wants ease by donating it’s loan pair electron to other atoms near to it. And Si has vacant d-orbital, so needy for it. Since both conditions are follow here,  hence back bonding will take place.

 Related Questions:

(1) Why trimethylamine amine ( N(CH3)3) is tetrahedral while trisilyl amine (N(SiH3)3) planner.?

(2) Why trimethyl amine {(CH3)3 N:} is pyramidal while trisilyl amine {(SiH3)3N:} is trigonal planer?

(3) Which is/are the correct statement/s about structure of methyl isothiocyanate (H3CNCS) and Silyl isothiocyanate(SiH3NCO)?

(4) What is the Si–N–C bond angle in Silyl isothiocyanate and methyl isothiocyanate (H3CNCS)?

(5) What are the order of extent back bonding, Lewis acid character and nucleophilicity of (BF3, BCl3, BBr3, BI3)boron trihalides?

(6) Arrange the silicon halides into decreasing order of Lewis acids Character? SiF4, SiCl4, SiBr4, SiI4

(7) Which of the following correct statement about structure of Ortho boric acid (H3BO3 ) is/are? Statements are as

(8) Which of the following is correct order of (B-O) bond length of following compounds? (1) B(OH)3 (2) B(OMe)3 and (3) B(Me)2OH

(9) What are the structure of (1) N(SiH3)3 (2) {(Me)3Si}2N-Si(Me)3 (3) HN(SiH3)2 and give the Answers of following questions?