Tutor profile: Borna Z.

How many unpaired electrons are there in $$[_{28}Ni(H_2O)_6]^{2+}$$?

In this question, we have to figure out the electron configuration of the metal center (Ni) first, and then count the number of unpaired electrons in the complex. Since, the ligands are all neutral ($$H_2O$$), the charge on Ni is 2+. Based on the aufbau principle, the electron configuration of Ni would be: [$$_{28}$$Ni] = 1s$$^2$$2s$$^2$$2p$$^6$$3s$$^2$$3p$$^6$$3d$$^8$$4s$$^2$$ Now we need to remove 2 electrons from the outermost layer which is 4s in this case. Hence, [$$_{28}$$Ni]$$^{2+}$$ = 1s$$^2$$2s$$^2$$2p$$^6$$3s$$^2$$3p$$^6$$3d$$^8$$ As can be seen, all electrons are paired up except in 3d orbital. There are 5 orbitals in 3d each can have up to 2 electrons. According to the Hund's rule we have to first fill all empty orbitals and then pair up the electrons if we have any additional electrons. So the first 5 electrons would fill the orbitals as _$$\uparrow $$_ _$$\uparrow $$_ _$$\uparrow $$_ _$$\uparrow $$_ _$$\uparrow $$_ and the remaining 3 electrons would pair up with the first 3 electrons in the first 3 orbitals leaving 2 unpaired electrons in the system: _$$\uparrow \downarrow $$_ _$$\uparrow \downarrow$$_ _$$\uparrow \downarrow$$_ _$$\uparrow $$_ _$$\uparrow $$_ Hence, there are 2 unpaired electrons in the system.

Under reversible conditions, derive an expression for the work done on an ideal gas in an isothermal process.

Based on the First law of thermodynamics, the work done on the system can be written as $$ w = -\int{P_{ex}dV}$$ where $$P_{ex}$$ is the external pressure. Since the process is reversible we have $$P_{ex}=P_{in}$$ and based on the ideal gas law we can substitute $$P_{in}$$ with $$nRT/V$$. As a result, we can write down work as $$ w = -\int{nRT/VdV}$$. The process is isothermal meaning that the T is constant; therefore, $$ w = -nRT\int{1/VdV}=-nRTln(V_2/V_1)$$.

What is the electron configuration of $$_{25}$$Mn?

Based on the aufbau principle, electrons fill atomic orbitals with the lowest energy levels first, i.e, orbitals with lower $$n+l$$ value. Hence, the order of filling would be: 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc... Note that 4s is filled before 3d since it has a lower $$n+l$$ value (4+0 < 3+2). Therefore, we have [$$_{25}$$Mn]: 1s$$^2$$2s$$^2$$2p$$^6$$3s$$^2$$3p$$^6$$4s$$^2$$3d$$^5$$.