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Eddi Din [679]
3 years ago
11

Consider the SCl2 molecule. (a) What is the electron config- uration of an isolated S atom? (b) What is the electron con- figura

tion of an isolated Cl atom? (c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl2? (d) What valence orbitals, if any, remain unhybrid- ized on the S atom in SCl2?
Chemistry
1 answer:
iragen [17]3 years ago
7 0

Answer:

(a) 1s² 2s² 2p⁶ 3s² 3p⁴

(b) 1s² 2s² 2p⁶ 3s² 3p⁵

(c) sp³

(d) No valence orbital remains unhybridized.

Explanation:

<em>Consider the SCl₂ molecule. </em>

<em>(a) What is the electron configuration of an isolated S atom? </em>

S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.

<em>(b) What is the electron configuration of an isolated Cl atom? </em>

Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.

<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>

SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.

<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>

No valence orbital remains unhybridized.

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Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
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For the given chemical reaction:

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]

Taking the standard enthalpy of formation:

\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of B_5H_9 produces -9078.57 kJ of energy.

Or,

If (2\times 63.12)g of B_5H_9 produces -9078.57 kJ of energy

Then, 1 gram of B_5H_9 will produce = \frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ of energy.

Hence, the amount of energy released per gram of B_5H_9 is -71.92 kJ

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