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2 years ago

Which of the following trends is indirectly proportional to effective nuclear charge, Zeff

Chemistry

2 answers:

Julli [10]2 years ago

7 0

Answer : Option C) Atomic Size

Explanation : The atomic radius of the elements is found to be decreasing if we go from left to right in the modern periodic table. Accordingly, $Z_{eff}$ increases as the number of shielding electrons present in the atomic nucleus of the periodic elements which lies in the same row remains constant while the number of protons in each atomic shell increases.

The effective nuclear charge $Z_{eff}$ of an atom is defined as the net positive charge which is felt by the valence electron of the atomic element.

When $Z_{eff}$ is observed to decrease, it is seen that the atomic radius grows in size. So, it explains the inverse relationship between both. This phenomenon occurs, because there is more screening of the electrons from the nucleus taking place, which is observed due to decrease the attraction between the electron and the nucleus.

ra1l [238]2 years ago

3 0

Periodic trends provides information about an element’s properties. One major trends in the periodic table is the atomic size, which is indirectly proportional to effective nuclear charge. The atomic size gradually decreases from left to right across a periodic table because all electrons are added to the same shell. At the same time, protons are also added to the nucleus, making it more positively charged.

Due to this, there is a greater nuclear attraction; the effect of increasing proton number is greater than that of the increasing electron number, which means that there is a strong attraction of nucleus manifested by pulling the atom's shell closer to the nucleus. Then, the valence electrons are seized closer towards the nucleus of the atom, which resulted to the decrease in atomic radius.

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Looking at the same nonmetal group on the periodic table, how does the reactivity of an element in period 2 compare to the react

Bas_tet [7]

Answer : Option B) The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.

Explanation : The reactivity of the Periods decreases as we go from left to right across a period. The farther to the left and down the periodic chart we go, the easier it is for electrons to be donated or taken away, resulting in higher reactivities of the elements. The attractive force of the protons is found to be stronger when electrons are found to be attracted to a closer electron shell.

4 0

2 years ago

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From the following reaction and data, find (a) S o of SOCl2 (b) T at which the reaction becomes nonspontaneous SO3(g) + SCl2(l)

disa [49]

Answer:

618 J/Kmol

T > 1.36 x 10³ K

Explanation:

The balanced reaction of interest is:

SO₃ (g) + SCl₂ (l) ⇒ SOCl₂ (l) + SO₂ (g)

with the data:

ΔHºf (kJ/mol) -396 -50.0 -245.6 -296.8

Sº(J/mol·K) 256.7 184 ? -248.1

ΔGº= -75.2 kJ

We know, we can find the standard change inGibb´s free energy from the equation:

ΔGºrxn = ΔHºrxn - TΔSºrxn

So we can calculate ΔHºrxn = ∑ ΔHºf prod - ΔHºreact, and substitute into this equation to solve Sº SOCl₂.

ΔHºrxn = ( -245.6 + (-296.8) ) - ( -396 - 50) kJ = - 96.4 kJ

Similarly for ΔSºrxn

ΔSºrxn = (-0.248.1 +Sº SOCl₂) - (0.256.7 +0.184) kJ/K

= -0.689 kJ /K -+ Sº SOCl₂

Plugging the values for the expression for ΔGºrxn:

-75.2 kJ = -96.4 kJ - 298 K x ( -0.689 kJ /K + Sº SCl₂ )

-75.2 kJ = -96.4 kJ + 205.3 Kj - 298 Sº SCl₂

-184 kJ = -298 K x Sº SCl₂

0.618 kJ/molK = Sº SCl₂

= 0.618 kJ/K x 1000 J = 618 J/Kmol

For the second part we will still be using the Gibb´s free energy change equation as above , but now we will solve for T when the reaction becomes non-spontaneous.

For the reaction to become non-spontaneous ΔGº is positive, and this happens when the term TΔSº becomes greater tha ΔHº:

ΔGºrxn = ΔHºrxn - TΔSºrxn

0 = ΔHºrxn - TΔSºrxn ⇒ TΔSºrxn = ΔHºrxn

T= ΔHºrxn / ΔSºrxn

ΔSºrxn = -0.689 J/Kmol + 0.618 J/Kmol = -0.0710 kJ/Kmol

( using the value the value just calculated from above )

T = - 96.4 kJ / -0.071 kJ/K = 1.36 x 10³ K

For temperatures greater than 1.36 x 10³ K the reaction becomes non-spontaneous.

4 0

2 years ago

Iodine has a lower atomic weight than tellurium (126.90 for I, 127.60 for Te) even though it has a higher atomic number (53 for

andrew11 [14]

Answer and Explanation:

Iodine have lower atomic mass than tellurium even though the atomic number of iodine is more than the atomic number of tellurium

This is because the atomic weight of any element is the sum of number of proton and number of neutron, even though the number of proton in iodine is more so but the number of neutron is less as compared to tellurium which makes the tellurium of high atomic mass

6 0

2 years ago

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

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jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f

loris [4]

Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

$H_2O(s)\rightleftharpoons H_2O(l)$

8 0

2 years ago

Read 2 more answers