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During the change of O_{2} to O_{2}^{-}, the incoming electron goes to the orbital :

 

  • Option 1)

    \pi 2 p_{y}

  • Option 2)

    \sigma ^{*}2p_{z}

  • Option 3)

    \pi^{*}2p_{x}

  • Option 4)

    \pi 2 p_{x}

For  molecule , the molecule orbital sequence  follows the given order :       =    =  Now, from  to  , the incoming electron will go into () orbital and thus  the final sequence will be        =    =   option (3) is correct.Option 1)Option 2)Option 3)Option 4)

The correct statements among I to III are  :

(I) Valence bond theory cannot explain the color exhibited by transition metal complexes .

(II)  Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.

(III) Valence bond theory cannot distinguish ligands as weak and strong field ones.

  • Option 1)

    II and III only 

  • Option 2)

    I  , II and III

  • Option 3)

    I and III only 

  • Option 4)

    I and II only 

  Valence bond theory ( VBT ) - 1.  It is formed by overlapping of valence shell atomic orbital of the two atoms having unpaired electron. 2.  There is maximum electron density between the bonding atoms 3.  Greater the overlapping of atomic orbital higher is the strength of chemical bond  - wherein     Limitation of Valence bond Theory - If does not explain 1.  Formation of chemical bond of...

Among the following species  , the diamagnetic molecule is :

  • Option 1)

    NO

  • Option 2)

    CO

  • Option 3)

    B_{2}

  • Option 4)

    O_{2}

  Energy level diagram for molecular orbitals. - 1S atomic orbitals combine to give    and   2S and 2P atomic orbitals give rise to eight molecular orbitals. - wherein     Magnetic behaviour of molecule - If  all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic - wherein However if one or more molecular orbital are singly occupied it is paramagnetic e.g....

HF has the highest boiling point among hydrogen halides because it has :

  • Option 1)

    Strongest vander waals interaction 

  • Option 2)

    Lowest ionic character 

  • Option 3)

    Strongest hydrogen bonding

  • Option 4)

    Lowest dissociation enthalpy

  Hydrogen Bonding - It is an electrostatic attractive force between hydrogen of one molecule or part of an molecule and an electronegative atom ( F, O, N ) of another molecule (intermolecular ) or another part of same molecule ( intramolecule ) - wherein     Hydrogen Bonding - Attractive force which binds hydrogen atom of one molecule with the electronegative ( F, O, N ) atom of another...

Among the following, the molecule expected to be stabilized by anion formation is :

C_{2},O_{2},NO,F_{2}

  • Option 1)

     C_{2}   

  • Option 2)

    F_{2}

  • Option 3)

    NO

  • Option 4)

    O_{2}

we know , stability of molecules  Bond order . In  , the  enters in  orbital So the bond order  will increase by  So, the stability would increase.Option 1)    Option 2)Option 3)Option 4)

The ion that has sp3d2 hybridization for the central atom is:

  • Option 1)

    [ICl4]-

  • Option 2)

    [ICI2]-

  • Option 3)

    [IF6]-

  • Option 4)

    [BrF2]-

[ICI4]- Steric Number formula:- S= No. of monovalent surrounding atom V=No. of valent electon of the central atom C=Charge on the molecule                                          If steric No=6 then hybridisation=sp3d2 so, ans is  [ICI4]-Option 1)[ICl4]-Option 2)[ICI2]-Option 3)[IF6]-Option 4)[BrF2]-

The correct statement about ICl_{5} and ICl_{4}^{-} is:

  • Option 1)

    both is isostructural

  • Option 2)

    ICI_{5} is trigonal bipyramidal and ICI_{4}^{-} is tetrahedral

  • Option 3)

    ICI_{5} is square bipyramidal and ICI_{4}^{-} is tetrahedral

  • Option 4)

    ICI_{5} is square bipyramidal and ICI_{4}^{-} is square planar.

 is square bipyramidal and  is square planar.   :-      Square pyramidal  l.P=1 b.P=5 hybridisation=     :-  Square planar l.p=2  b.p=4 hybridisation=  Option 1)both is isostructuralOption 2) is trigonal bipyramidal and  is tetrahedralOption 3) is square bipyramidal and  is tetrahedralOption 4) is square bipyramidal and  is square planar.

Among the following molecules/ions, C_{2}^{2-}, N_{2}^{2-}, O_{2}^{2-},O_{2} which one is diamagnetic and has the shortest bond length?

  • Option 1)

    O2

  • Option 2)

    N_{2}^{2-}

  • Option 3)

    O_{2}^{2-}

  • Option 4)

    C_{2}^{2-}

:- No of electrons = 14    (1) It is diamagnetic (2) B.O=    :- No of electrons=18     (1) It is diamagnetic (2) B.O=     ans is Option 1)O2Option 2)Option 3)Option 4)
As options are not given in the questions, so I am solving this question considering these below options. You can relate to your approach in the same way. Options: Thus, among the given pairs, only  have  hybridisation.
As we learnt in Magnetic behaviour of molecule - If  all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic - wherein However if one or more molecular orbital are singly occupied it is paramagnetic e.g. O2 Electronic configuration of  molecule  Since there are 2 unpaired electrons in  &  degenerate orbitals, so,  is paramagnetic. Similarly, for  molecule,...
As we learnt in Bond Order - Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals. - wherein  We know that bond order  of all the species given,  has the highest bond order at 2.5 thereby giving it the lowest bond length. Option 1) This is the correct option Option 2) This is an incorrect option Option 3) This...

 

 

a lone pair refers to a pair of valence electrons that are not shared with another atom and is sometimes called an unshared pair or non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. They can be identified by using a Lewis structure.
Screenshot_1232.png Inbox (180) - nelantisanthoshku- X https://mail.goagle.com/mail/u/O/?pli=1#inbox?projector= 1 Revision Test Series for JEE (Main) - 2019 (Phase-II) Home Assignment-8 (Code-D-1) Which of the following has maximum total spin? 34. 35. 36. 37. 38. The ratio of root mean square speed of NO gas to most probable speed of C2H6 gas at 298 K is (1) 1.73 (3) 1.06 (2) 1.41 (4) 1.22 42. 43. 44. 45. 46. (3) Mn+2 (2) cr*2 (4) Cu*2 Contact process is used for manufacturing of What is the formula of crystal containing atom X at each corner of cube and Y at each edge centre? (l) X2Y Identify the property which is both extensive and state (1) HN03 (3) N02 Which allotrope chemiluminescence? (1) White (3) Black (2) H2S04 (4) H2S of phosphorus (2) Red (4) Blue show function. (1) Heat (3) Work (2) Enthalpy (4) Both (1) and (3) Two miscible liquids X and Y are mixed in 1 : 3 mole ratio. What is the total vapour pressure of mixture? (Vapour pressures of pure liquid X and Y are 300 mm and 200 mm respectively) What is the % p-character in PC15? (4) 66.67% Carbon cannot be used as reducing agent at 9000C 1625 Wednesday 27-02-2019 5 for (1) Feo (1) 210 mm (3) 225 mm (2) 275 mm (4) 250 (2) cup Activate Windows (4) ZnO Go to Settings to activate W n enthalpy is most negative for : ows. 6 19 The yield of following reaction is 50

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 sp3d2 hybridization is not displayed by :

  • Option 1)

     BrF5

  • Option 2)

     SF6

  • Option 3)

    [CrF6]3−

  • Option 4)

     PF5

     

 
Option 1)  BrF5 Option 2)  SF6 Option 3) [CrF6]3− Option 4)  PF5  

Which of the following is paramagnetic ?

  • Option 1)

    NO+

  • Option 2)

     CO

     

  • Option 3)

     O _{2}^{2-}

  • Option 4)

    B2

 
Option 1) NO+ Option 2)  CO   Option 3)   Option 4) B2
In H-O-H, there are two lone of electrons are present on the Oxygen atom which push the two O-H bonds towards each other and thus the bond angle in H-O-H reduces to  from . But in the case of , it is a regular geometry with no lone pairs present, hence the H-C-H bond angle is 

The type of hybridisation and number of lone pair (s) of electtrons ofXe \: \: \: in \: \: XeOF_{4} respectively , are :

  • Option 1)

    sp^{3}d^{2}\: \: and \: 1

  • Option 2)

    sp^{3}d\: \: and \: 2

     

  • Option 3)

    sp^{3}d^{2}\: \: and \: 2

  • Option 4)

     

    sp^{3}d\: \: and \: 1

  Hybridisation - The process of mixing of atomic orbitals belonging to the same atoms of slightly different energies so that a redistribution of energy takes place between them  resulting in the formation of new set of orbital of equivalent energies and shape is called hybridisation. - wherein The new orbitals thus formed are called hybrid orbitals         VSEPR Theory -   1.  The shape of the...

Two pi and half sigma bonds are present in :

  • Option 1)

    O_{2}^{+}

  • Option 2)

    N_{2}

  • Option 3)

    O_{2}

  • Option 4)

    N_{2}^{+}

  Pi bond - This type of bond is formed by sidewise or lateral over lapping of two half filled atomic orbitals. - wherein              Pi bond - This type of bond is formed by sidewise or lateral over lapping of two half filled atomic orbitals. - wherein     Sigma bond - When covalent bond is formed by overlapping of atomic orbitals along the same axis it is called Sigma bond. Such type of...

In which of the following processes , the bond order has increased and paramagnetic charecter has changed to  diamagnetic ?

  • Option 1)

    N_{2}\rightarrow N_{2}\: ^{+}

  • Option 2)

    O_{2}\rightarrow O_{2}^{2-}

  • Option 3)

    O_{2}\rightarrow O_{2}\: ^{+}

  • Option 4)

    NO\rightarrow NO^{+}

    Bond Order -   Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals. - wherein         Magnetic behaviour of molecule - If  all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic - wherein However if one or more molecular orbital are singly occupied it is paramagnetic e.g....

According to molecular orbital theory, which of the following is true with respect toLi_{2}^{+}   and Li_{2}^{-}?

  • Option 1)

    Li_{2}^{+}is unstable and isLi_{2}^{-} stable

  • Option 2)

    Li_{2}^{+}is stable andLi_{2}^{-} is unstable

  • Option 3)

    Both are stable

  • Option 4)

    Both are unstable

  Molecular Orbital Theory ( M.O.T ) - Molecular orbital ( M.O ) Theory was developed by F Hund and  R.S Mulliken in 1932. -     Features of Molecular Orbital Theory - 1.  Bonding molecular orbital has lower energy and hence greater stability than corresponding antibonding molecular orbital. 2.  The electron probability distribution around a group of nuclei in a molecule is given by molecular...
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