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The bond angle in H(2) S is:...

The bond angle in `H_(2) S` is:

A

`gt NH_3`

B

Same as in `BeCl_2`

C

`gt H_(2)Se, ltH_2O`

D

Same as in `CH_4`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the bond angle in H₂S, we will analyze the molecular geometry and the effects of lone pairs on the bond angle. ### Step 1: Determine the molecular geometry of H₂S H₂S (hydrogen sulfide) has a central sulfur atom bonded to two hydrogen atoms. The sulfur atom has two lone pairs of electrons. The presence of these lone pairs affects the molecular geometry. **Hint:** Consider the VSEPR theory, which helps predict the shape of molecules based on the repulsion between electron pairs. ### Step 2: Apply the VSEPR theory According to VSEPR theory, the arrangement of electron pairs around the central atom determines the molecular shape. In H₂S, we have: - 2 bonding pairs (from H-S bonds) - 2 lone pairs This arrangement corresponds to a bent or angular shape. **Hint:** Remember that lone pairs occupy more space than bonding pairs, influencing the bond angles. ### Step 3: Analyze the bond angle In a perfect tetrahedral arrangement (which has a bond angle of 109.5°), the presence of lone pairs will compress the bond angle between the hydrogen atoms. In H₂S, the bond angle is reduced due to the lone pair-lone pair and lone pair-bond pair repulsion. **Hint:** Compare the bond angles in related molecules (like H₂O and NH₃) to understand how lone pairs affect the angles. ### Step 4: Compare with other molecules - In H₂O, the bond angle is approximately 104.5° due to two lone pairs. - In NH₃, the bond angle is approximately 107° due to one lone pair. Since sulfur is larger than oxygen, the bond angle in H₂S will be less than that in H₂O but greater than that in NH₃. **Hint:** Consider the trend in bond angles as you move down the group in the periodic table. ### Step 5: Conclusion The bond angle in H₂S is approximately 92.5°, which is less than that in H₂O and NH₃, but it is influenced by the size of the sulfur atom and the presence of lone pairs. **Final Answer:** The bond angle in H₂S is approximately 92.5°.
To determine the bond angle in H₂S, we will analyze the molecular geometry and the effects of lone pairs on the bond angle. ### Step 1: Determine the molecular geometry of H₂S H₂S (hydrogen sulfide) has a central sulfur atom bonded to two hydrogen atoms. The sulfur atom has two lone pairs of electrons. The presence of these lone pairs affects the molecular geometry. **Hint:** Consider the VSEPR theory, which helps predict the shape of molecules based on the repulsion between electron pairs. ### Step 2: Apply the VSEPR theory ...
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Explore conceptually related problems

The bond angle of H_(2)Se is best described as

Why H-S-H bond angle in H_(2)S is smaller than H-O-H bond angle in H_(2)O ?

Knowledge Check

  • The bond angle in H_2S is

    A
    `gt NH_3`
    B
    Same as in `BeCl_2`
    C
    `gt H_2Se`, `lt H_2O`
    D
    Same as in `CH_4`

  • Chemical bonding between two atoms is necessarily associated with an electrical moment arising out of the difference in electroegativity of two atoms. This means that every bond carries with it an electrical moment called the ''bond moment''. The dipole moment of a molecule is really the vectorial sum of the individual bond moment present in it. To compute the dipole moment it is necessary to find out the values of various bond moment. In the following table, dipole moment of different bonds are as given. {:(,rarr,rarr,rarr),("Bond",H-C,C-Cl,C=0),("Bond moments",0.4 D,1.5 D,2.5 D):} The group moments of few groups as given {:("Group",NO_(2),OH,CN,CH_(3)),("Direction of dipole","Towards N","Towards O","Towards N","Away from "CH_(3)),("Dipole moment",4D,1.6D,3.8D,0.4D):} The bond angle in H_(2)S is 97^(@) and its dipole moment is 1.5 D. The S-H bond distance is 0.15 nm. Therefore, approximate percentrage ionic cjaracter pf S-H bond is (neglect the effect of dipole moment of lone pair on sulphur atom in H_(2)S ). (Given : [cos 97^(@)=-0.121 and sqrt(0.88)=0.94] )

    A
    `32%`
    B
    `16%`
    C
    `84%`
    D
    `10%`

  • Bond angle in H_(2)O is

    A
    `104.5^(@)`
    B
    `120^(@)`
    C
    `109.5^(@)`
    D
    `107^(@)`

    • Similar Questions

    Explore conceptually related problems

    Chemical bonding between two atoms is necessarily with an electrical moment arising out of the difference in electronegativity of two atoms. This means that every bond carries with it an electrical moment called the bond moment.The dipole moment of a molecule is really the vectorial sum of the individual bond moment present in it. To compute the dipole moment it is necessary to find out the values of various bond moment in the moment of different bonds are as given. {:("Bond",oversetto(H-C), " ",oversetto(C-Cl)," ",oversetto(C=O)),("Bond moments",0.4 D ," ", 1.5 D, " ", 2.5D):} The group moments of few groups as given {:("Group",NO_2,OH,CN,CH_3),("direction of dipole","toward N","towards O","towards N","away from" CH_2),("Dipole moment",4D,1.6D,3.8 D,0.4D):} The bond angle in H_2S is 97^@ and its dipole moments is 1.5 D.The S-H bond distance is 0.15 nm.There fore approximate percentage ionic character of S-H bond is (neglect the effect of dipole moment of lone pair on sulphur atom in H_2S ).(Given [cos 97^@=-0.12 ] and sqrt0.88=0.94 )

    What is the H-S-H bond angle in H_(2)S ?

    The bond angle in H_2O is nearly 105^@ whereas bond angle in H_2 S is nearly 92^@ . This is because.

    Assertion . H-S-H bond angle in H_(2)S is closer to 90^(@) but H-O-H bond angle in H_(2)O is 104.5^(@) . Reason . Lp-lp repulsion us stronger ub H_(2)S than in H_(2)O

    Assertion : H-S-H bond angle in H_(2)S is closer to 90^(@) but H-O-H bond angle in H_(2)O is 104.5^(@) Reason: lp-lp repulsion is stronger in H_(2)S than in H_(2)O

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