Structure of Atom

Statement is FALSE. The outer electronic configuration of the ground state chromium atom is unusual compared to what might be expected based on a straightforward application of the Aufbau principle. The actual outer electronic configuration of chromium is $3d^{5} 4s^{1}$, not $3d^{4} 4s^{2}$.

This configuration arises because a half-filled d subshell ($3d^{5}$) is more stable than the configuration $3d^{4} 4s^{2}$. By promoting an electron from the 4s orbital to the 3d orbital to achieve a half-filled d subshell, chromium minimizes its energy. This is due to the increased exchange energy and symmetry of the half-filled d subshell, leading to greater stability. Thus, the correct outer electronic configuration of a ground state chromium atom is $3d^{5} 4s^{1}$.

Rutherford's experiment on the scattering of $\alpha$-particles, conducted between 1908 and 1913, was groundbreaking in the field of atomic physics. In this experiment, Rutherford and his colleagues, Hans Geiger and Ernest Marsden, aimed a beam of alpha particles (which are positively charged particles) at a thin gold foil and observed their scattering patterns using a fluorescence screen.

The results were surprising and led to a dramatic shift in the understanding of the atomic structure. Most of the alpha particles passed straight through the gold foil, a few were deflected at large angles, and a very small number were deflected backward. This was in stark contrast to the expectations based on the plum pudding model of the atom that was prevalent at the time, which proposed that the atom was a diffuse cloud of positive charge with electrons embedded throughout.

The only way Rutherford could explain the observed scattering patterns was by proposing that the atom consists of a tiny, dense nucleus that contains all of its positive charge and most of its mass. This nucleus is surrounded by a cloud of electrons, which occupies the rest of the atom's volume but contributes negligibly to its mass. The alpha particles were mostly passing through the empty space of the atom, but when they came close to the dense nucleus, they were deflected by its positive charge.

Therefore, Rutherford's experiment showed for the first time that the atom has a nucleus, making the correct answer:

Option C: nucleus.

To find the number of neutrons in a dipositive zinc ion (Zn$^{2+}$) with a mass number of 70, we first need to understand what each term means and then perform the calculation:

1. Mass number: This is the total number of protons and neutrons in an atom's nucleus. For this zinc ion, the mass number is given as 70.

2. Zinc's atomic number: The atomic number of an element represents the number of protons in the nucleus of an atom of that element. For zinc, the atomic number is 30, which means every zinc atom has 30 protons.

3. Dipositive zinc ion (Zn$^{2+}$): This indicates a zinc ion that has lost two electrons, becoming positively charged. However, the loss of electrons does not affect the number of protons or neutrons in the nucleus, so the atomic and mass numbers are unaffected by the ion's charge state.

To find the number of neutrons in the zinc ion, we subtract the number of protons (atomic number) from the mass number:

$ \text{Number of neutrons} = \text{Mass number} - \text{Atomic number} $

For the zinc ion in question:

$ \text{Number of neutrons} = 70 - 30 = 40 $

Therefore, the number of neutrons in the dipositive zinc ion with a mass number of 70 is 40, which corresponds to option D.