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3 years ago

What is the direction of the polarity of the indicated bond in h3c−oh marked by δ+ and δ−?

Chemistry

1 answer:

Sedbober [7]3 years ago

6 0

Answer:- The direction of the polarity of the indicated bond is from carbon to oxygen.

Explanations:- There are two types of covalent compounds, polar and non polar. If the bond is between two same atoms for example, H-H, Cl-Cl etc then the bond is non polar. If the bond is between two different atoms then the bond would be polar. The direction of the polarity is from loss electron negative atom to more electron negative atom.

Oxygen is more electron negative than carbon. So, being more electron negative, the bonding electrons are more towards oxygen and it cases partial negative charge on oxygen and partial positive charge on carbon. The direction of the polarity is from less electron negative carbon to more electron negative oxygen.

It is shown in the diagram below:

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

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