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2 years ago

based on what you wrote in the table in part a, what effect do lone pairs have on the bond angle? why do you think this happens?

1 answer:

6 0

Because they are more in close proximity to the core atom's nucleus than lone pairs are, lone pairs reject other lone pairs more strongly than bonding pairs do effect the bond angle.

When a single pair of electrons at the central atom begins to resist the bound pair of electrons, the bond angle decreases and the bonds are slightly shifted inward. It moves existing atoms closer together and modifies their geometry. It's equivalent to include an atom. When there is an increase in back bonding, the bond angle rises. Because the lone pair electrons of the two atoms reject one another, adjacent atoms in a molecule that have lone pair electrons will not be kept together in their bond as securely.

learn more about bond angle here:

brainly.com/question/26587488

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In the following reaction, which is a Brønsted-Lowry base?

Mariulka [41]

Answer:

C2O42–(aq)

Explanation:

This accepted hydrogen from H3O (hydronium) so it is a Bronsted-Lowry Base.

7 0

2 years ago

Any help welcome on this, chem balancing equations btw !

Montano1993 [528]

Answer:

A. 1, 1, 1, 1

B.1,1,1

C. 2, 1, 2, 1

D. 2, 1, 1, 1

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7 0

3 years ago

Brainliest get's +20 points!

gizmo_the_mogwai [7]

Answer:

i think thats correcr!

Explanation:

6 0

4 years ago

How many total ions are in 2.95 g of magnesium sulfite?

Aleksandr [31]

Answer:

Approximately $5.65 \times 10^{-2}\; \rm mol$ , which is approximately $3.41 \times 10^{22}$ particles.

Explanation:

Formula of magnesium sulfite: $\rm MgSO_3$ .

Look up the relative atomic mass of $\rm Mg$ . $\rm S$ , and $\rm O$ on a modern periodic table:

$\rm Mg$ : $24.305$ .
$\rm S$ : $32.06$ .
$\rm O$ : $15.999$ .

The ionic compound $\rm MgSO_3$ consist of magnesium ions $\rm {Mg}^{2+}$ and sulfite ions $\rm {SO_3}^{2-}$ .

Notice that $\rm {Mg}^{2+}\!$ and $\rm {SO_3}^{2-}\!$ combine at a one-to-one ratio to form the neural compound $\rm MgSO_3$ . Therefore, each $\rm MgSO_3\!$ formula unit would include one $\rm {Mg}^{2+}$ ion and one $\rm {SO_3}^{2-}$ ion (that would be two ions in total).

Calculate the formula mass of one such formula unit:

$\begin{aligned}&\; M(\mathrm{MgSO_3}) \\ = & \; 24.305 + 32.06 + 3 \times 15.999 \\ = & \; 104.362\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the mass of one mole of $\rm MgSO_3$ formula units (which includes one mole of $\rm {Mg}^{2+}$ ions and one mole of $\rm {SO_3}^{2-}$ ions) would be $104.362\; \rm g$ .

Calculate the number of moles of such formula units in that $2.95\; \rm g$ of $\rm MgSO_3$ :

$\begin{aligned}n&= \frac{m}{M}\\ &=\frac{2.95\; \rm g}{104.362\; \rm g \cdot mol^{-1}} \approx 2.82670 \times 10^{-2}\; \rm mol\end{aligned}$ .

There are two moles of ions in each mole of $\rm MgSO_3$ formula units. Therefore, that $2.82670 \times 10^{-2}\; \rm mol$ of $\rm MgSO_3\!$ formula units would include approximately $2 \times 2.82670 \times 10^{-2}\; \rm mol \approx 5.65\times 10^{-2}\; \rm mol$ of ions.

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3 years ago

Give the electron geometry (eg), molecular geometry (mg), and hybridization for xef4. give the electron geometry (eg), molecular

Ostrovityanka [42]

Hello!

answer : c

eg = octahedral, mg = square planar, sp3d2

Hope that helps!

6 0

3 years ago