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?

Penetration Effect: Penetration Power:

The probability of an electron which allows an electron to get close to the nucleus is known as penetration effect. It is known as the proximity of the electron in the orbital to the nucleus.

(1)  The relative penetration power of sub shells within same shell (same value of n) follow the order as:

                                        s>p>d>f

We consider it for each shell and sub shell as the relative density of the electrons near the nucleus of atoms. It is clear that the ‘S’ electrons have greater probability of coming closer to the nucleus than the P, d or f electrons of the same principal energy shell. 

(2) For different values of shell (n) and sub shell (l), decreasing penetrating power of an electron follows as:

             1s>2s>2p>3s>3p>4s>3d>4p>5s>4d>5p>6s>4f....

(3) In other words, ‘S’ electrons penetrate (more nearer) more towards the nucleus than the ‘P’ electrons and the penetrating power of the electrons in the given principal energy shell varies as S> P> d>f. Thus the ‘S’ electrons experience more attraction from the nucleus than the p d or f electrons of the same principal energy shell.

Therefore, greater energy required to remove out electron from‘s’ orbital than ‘p’, d and ‘f’ orbital. Thus the ionization potential for pulling out an ‘s’ electron is maximum  and it decreases in pulling out a p , d or f electron of the same principal energy shell.

(4) Ionization potential or Ionization enthalpy of an atom is directly proportional to penetration power of orbitals.

Application of Penetration effect:

Example: Ionization energy of Boron is smaller than Beryllium even though effective nuclear charge is higher?

Solution: The electronic configurations of Boron and Beryllium are (₅B=1S²,2S²,2p¹) and (₄Be =1S²,2S²).

In Boron the outermost electron is present in the 2p orbital (low penetration power) and is less strongly bound than the electron present in a 2S orbital of Beryllium(Have more penetration power), which will has a higher Zeff. It is easier to ionize the Boron atom.

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 ?