Chemistry, 01.08.2020 01:01 mparra4761

Which one of the following is most likely to gain electrons when forming an ion, based on the natural tendency of the element? A Ni
B S
C Na
D Cr
E Be

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Answer from: xjaliw11882

Answer

B) Sulphur (S)

Explanation

Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.

Now, let us understand what is a metal and a non-metal.

Metal- electron donors are called as metal.Non-metal- electron acceptors are called non-metals.

So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.

Now let us also look at the electronic configuration of Sulphur to get the picture more clearly

atomic no. of sulphur would be = 16

$S\rightarrow 1s^2\; 2s^2\;2p^6\;3s^2\;3p^4$

so here the p-subshell is incomplete and is in need of 2 electrons.

Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.

To learn more about Ion Formation

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Answer from: aallyssabrown0120

Option B. S

Explanation:

All of the options except sulphur, S is metal.

Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.

Sulphur, S has atomic number of 16 with electronic configuration as:

S (16) => 1s² 2s²2p⁶ 3s²3p⁴

From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.

Therefore, sulphur, S will gain two electrons to form ion.

As stated earlier, the rest option given are all metals which will form ion by losing electron(s).

Answer from: Quest

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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