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deff fn [24]
4 years ago
11

Based on VSEPR theory and your observations from the Molecular Geometry lab consider the following questions What is the predict

ed hybridization at an atom which is surrounded by a double bond and two single bonds?
a) Sp
b) sp^2
c) sp^3
Chemistry
1 answer:
____ [38]4 years ago
6 0

Answer:

b) sp^2

Explanation:

Hybridization refers to the concept that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. In chemistry, orbital hybridisation (or hybridization) is the implies the mixing of atomic orbitals to form hybrid orbitals (with different energies, shapes, etc., different from that of the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds according to the principles of the valence bond theory.

In 1931 Linus Pauling proposed the idea of “mixing” the orbitals or “hybridizing” them to account for certain observed bonding patterns. Pauling proposed a sort of a combination of the orbitals giving you an orbital that has partial characters.

Hybridization is merely a mathematical construct. It is never an actual “process” that occurs within orbitals . Hybridization is a mathematical model that describes how the atomic orbitals would’ve looked like based on the observable molecular orbitals.

sp2 hybridization leads to the formation of a double bond. sigma bonds may also be formed depending on the valency of the central atom. In alkenes, an sp2 hybridized carbon atom forms a double bond in addition to two sigma bonds to other atoms.

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<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

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