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In the process , O2 ^+ gives O2^+2 + e the electron lost is from (a0 Bonding pie orbital (b) antibonding pie orbital (c) 2pz orbital (d) 2px orbital
08
Dec
In the process , O2 ^+ gives O2^+2 + e the electron lost is from (a0 Bonding pie orbital (b) antibonding pie orbital (c) 2pz orbital (d) 2px orbital In the process O2 ^+ gives O2^+2 + e the electron lost is from (a0 Bonding pie orbital (b) antibonding pie orbital (c) 2pz orbital (d) [...]
The paramagnetic behaviour of B2 is due to the presence of (a) 2 unpaired electrons in pie b MO (b) 2 unpaired electrons in pie * MO
08
Dec
The paramagnetic behaviour of B2 is due to the presence of (a) 2 unpaired electrons in pie b MO (b) 2 unpaired electrons in pie * MO The paramagnetic behaviour of B2 is due to the presence of (a) 2 unpaired electrons in pie b MO (b) 2 unpaired electrons in pie * MO December [...]
Heterolytic bond association energy of alkyl halides follow the sequence (a) R-F > R-Cl > R-Br > R-I
08
Dec
Heterolytic bond association energy of alkyl halides follow the sequence (a) R-F > R-Cl > R-Br > R-I Heterolytic bond association energy of alkyl halides follow the sequence (a) R-F > R-Cl > R-Br > R-I December 8, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
The bond order in O2^- ion is (a) 2 (b) 1 (c) 2.5 (d) 1.5
08
Dec
The bond order in O2^- ion is (a) 2 (b) 1 (c) 2.5 (d) 1.5 The bond order in O2^- ion is (a) 2 (b) 1 (c) 2.5 (d) 1.5 December 8, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Consider the following compounds (a) Chloroethene B. Benzene C. 1,3 butadiene D. 1, 3, 5 hexatriene All the carbon atoms are sp2 hybridised in
08
Dec
Consider the following compounds (a) Chloroethene B. Benzene C. 1,3 butadiene D. 1, 3, 5 hexatriene All the carbon atoms are sp2 hybridised in 3 butadiene D. 1 3% 5 hexatriene All the carbon atoms are sp2 hybridised in Consider the following compounds (a) Chloroethene B. Benzene C. 1 December 8, 2020 Category: Uncategorised (JEE [...]
Number of pie electrons present in naphthalene is (a) 4 (b) 6 (c) 10 (d) 14
08
Dec
Number of pie electrons present in naphthalene is (a) 4 (b) 6 (c) 10 (d) 14 Number of pie electrons present in naphthalene is (a) 4 (b) 6 (c) 10 (d) 14 December 8, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Example of sp2 hybridization is (a) CH3 + (b) CH3 (c) C2H5 + (d) C2H5
08
Dec
Example of sp2 hybridization is (a) CH3 + (b) CH3 (c) C2H5 + (d) C2H5 Example of sp2 hybridization is (a) CH3 + (b) CH3 (c) C2H5 + (d) C2H5 December 8, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Toluene has (a) 6 sigma and 3 pie bond (b) 9 sigma and 3 pie bond (c) 9 sigma and 6 pie bond
08
Dec
Toluene has (a) 6 sigma and 3 pie bond (b) 9 sigma and 3 pie bond (c) 9 sigma and 6 pie bond Toluene has (a) 6 sigma and 3 pie bond (b) 9 sigma and 3 pie bond (c) 9 sigma and 6 pie bond December 8, 2020 Category: Uncategorised (JEE Advanced Physics by [...]
Hybridization of 1 and 2 carbon atoms in CH2 = C = CH2 (a) sp, sp (b) sp2, sp2 (c) sp2, sp
08
Dec
Hybridization of 1 and 2 carbon atoms in CH2 = C = CH2 (a) sp, sp (b) sp2, sp2 (c) sp2, sp Hybridization of 1 and 2 carbon atoms in CH2 = C = CH2 (a) sp sp sp (b) sp2 sp2 (c) sp2 December 8, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma [...]
The correct statement with regard to H^+2 and H^−2 is (a) Both H+2 and H−2 do not exist (b) H−2 is more stable than H+2 (c) H−2 is more stable than H+2 (d) Both H+2 and H−2 are equally stable
08
Dec
The correct statement with regard to H^+2 and H^−2 is (a) Both H+2 and H−2 do not exist (b) H−2 is more stable than H+2 (c) H−2 is more stable than H+2 (d) Both H+2 and H−2 are equally stable The correct statement with regard to H^+2 and H^−2 is (a) Both H+2 and H−2 [...]