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Keeping surrounding atom same if electronegativity of central atom increases bond angle increases.
According to Drag’s rule six molecules (PH3, AsH3, SbH3 H2S, H2Se and H2Te) are no Hybridisation having on their central atom.
If central atom is in third row of or below in periodic table then lone pair will occupy in stereo-chemically inactive S-orbital and bonding will be through p-orbital and near 900 angles. If electronegativity of surrounding atom is less than or equal < 2.5.
Hence (theta) is bond angle between equivalent orbital % S and P-character is given by S and P
1- It is applicable of them is % S + % P = 100%
2- All the orbital must not be involve
3- It is applicable when
About 600 k j mole-1 more energy is required so hybridized the central phosphorous (P) atom. due to this energy factor more stable arrangement is, when pure orbital’s are involved in bonding and lone pair present into pure s-orbital.
Q:1 Explain why NH3 is more soluble in water than PH3?
Q:2 The formation of PH4+ is very difficult but the formation of NH4+ is easier than PH4+ ?
Q:3 NH3 is stronger Lewis base  in comparison PH3?
Q:4 The complex formation Tendency of NH3 is wisher than PH3.
SOLUTION: In NH3 lane pair is present in one of the sp3 Hybridized orbital. While in PH3 lone pair is present in pure S orbital and hence lane pair donation capacity of NH3 is stronger than PH3
Paragraph for question nos. 5 & 6:
Drago suggested an empirical rule which is compatible with the energetic of hybridisation. It states that if the central atom is in the third row or below in the periodic table, the lone pair will occupy a stereo chemically inactive s-orbital, and the bonding will be through p-orbitals and bond angles with be nearly 90º if the electronegativity of the surrounding atom is  2.5.
Question (5): In which of the following molecule central atom has higher % s-character in its bond pair-
(A) AsH3                    (B) GeH4                  (C) P4                 (D) H2Se
Question (6): Correct order of bond angle is-
(A) PH4 + > OF2 > SF2> SbH3 > H2Te     (B) OF2 > SF2 > PH4 > SbH3 > H2Te
(C) PH4 + > SF2 > OF2 > SbH3 > H2Te    (D) SF2 > OF2 > PH4 + > SbH3 > H2Te

Paragraph for question nos. 7 & 8:
The more ionic character (i.e. poorer covalency) in a bond (due to the more electronegative substituent to the central atom) leads to the utilization of hybrid orbitals containing more p-character (i.e. less s-character) of the central atom. In other words, the multiple bonds encourage the central atom to utilize its hybrid orbitals which contains more s-character.
The percentage of s- or p-character between two adjacent and equivalent hybrid orbitals can determine from the knowledge of the corresponding bond angle () as follows:
Question (7): If x1, x2 and x3 are S – S bond lengths in S2O4-2, S2O5-2 and S2O6-2 respectively, then correct order for S – S bond length is -
(A) x3 > x2 > x1          (B) x1 > x2 > x3          (C) x3 > x1 > x2        (D) x1 > x3 > x2
Question (8): The percentage of p-character (approx) in hybrid orbital having the lone pair at central atom in SF4 molecule will be:
 (Given F S- F equatorial bond angle is 102º and cos102º = – 0.21)
(A) 65.5                    (B) 34.4                     (C) 50.5                        (D) 82.6
Paragraph for question nos. 9 & 10:
In all expected compounds each case central atom only uses its s and p orbitals in hybridisation. The relation between bond angle theta and decimal fraction of s and p character present in the equivalent hybrid orbitals is given by:
Here theta is bond angle between equivalent hybrid orbital (S and P)
Question (9): The correct order of % P character in bond pairs of central atoms in the following compounds:
(A) P>T>S>Q>R      (B) S>R>T>P>Q   (C) P>Q>S>R>T   (D) P>Q>S>T>R
Question (10): If Value of n is 2 for compound T, then number of lone pair present at central atom of compound T will be:
(A) 0                             (B) 1                              (C) 2                           (D) 3
Question (11): The correct statement is :
(A) The ratio of % p character to % s character is less than four, for the bond pair of central atom of compound S  
 (B) Central atom uses three hybrid orbitals to form compound R
 (C) Central atom uses four hybrid orbitals to form compound S
 (D) There are three compounds present between point C to E. according to % s character  in bond pair of central atom.

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