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1) sp – Hybridisation. In such hybridisation one s- and one p-orbital are mixed to form two sp – hybrid orbitals, having a linear structure with bond angle 180 degrees. For example in the formation of BeCl 2, first be atom comes in excited state 2s 1 2p 1, then hybridized to form two sp – hybrid orbitals. These hybrid orbitals overlap ...
What is the state of hybridisation of carbon in (a) CO2− 3 (b) diamond (c) graphite? View Solution. Q 2.
The hybridisation of atomic orbital of Nitrogen in N O− 3,N O+ 2,N H + 4 respectively: View Solution. Q 5.
The hybridisation of BeCl in solid state and above 1200 K is respectively1) sp3, sp3(3) sp2 sp2(2) sp3, sp2(4) sp3 , sp
What are hybridisation states of each carbon atom in the following compounds? C H 2 = C = O , C H 3 C H = C H 2 , ( C H 3 ) 2 C O , C H 2 = C H C N , C 6 H 6 View Solution
The empty orbital forms co-ordinate covalent with −OH − group. The boron atom in B(OH)3 is electron deficient: it has 6 rather than 8 electrons in its valence shell. It acts as a Lewis acid by readily accepting the lone pair from the oxygen in a water molecule to go from sp2 to sp3 hybridisation. Was this answer helpful?
The correct order of hybridisation of the central atom in the following species N H 3, [P t C l 4] 2 −, P C l 5 and B C l 3. View Solution. Q5. N i C l 2 + K C N ...
The hybridisation of atomic orbital of Nitrogen in N O − 3, N O + 2, N H + 4 respectively: View Solution ...
Explanation: 1. Carbon is the central atom in both C H 3 + as well as C H 3 −. herefore the hybridization of carbon is the hybridization of both the given molecules. 2. For a carbon atom, we have 4 valence electrons. 3. In the molecule C H 3 +, the positive charge represents removal of a electron. Therefore there are 3 valence electrons ...
Q 4. Find the magnitude of the standard electrode potential of M nO− 4/M nO2. The standard electrode potentials of M nO− 4/M n2+ =1.51 V and M nO2/M n2+ = 1.23 V. View Solution. Q 5. Oxidation state of Mn in M nO− 4 is +7 indicating all electrons paired in Mn but M nO− 4 is coloured. It is due to.
