Answer Explanation:
       
In the photoelectric effect, electrons in the outermost shell (highest energy level) are ejected first and have the highest kinetic energy because they are held less tightly by the nucleus. For sulfur, the electron configuration is:
        

1s2 2s2 2p6 3s2 3p4
        
The outermost electrons are in the 3p subshell.
        

Correct Answer: (C) 3p

 
Answer Explanation:
        
Energy of a photon is inversely proportional to its wavelength and directly proportional to its frequency:
        

            $$ E = \dfrac{hc}{\lambda} $$
        
        
Infrared radiation has a longer wavelength (\(10^{-5}\) m) and therefore a lower frequency and less energy compared to ultraviolet radiation.
        
Correct Answer: (A) Infrared, because it has a lower frequency.

 
Answer Explanation:
        
The PES spectrum shows peaks corresponding to the electron configuration of phosphorus:
        

            $$ \text{P}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^3 $$
        
        
The peaks correspond to the binding energies of the electrons in different subshells. Phosphorus has 15 electrons, which matches the spectrum.
        
Correct Answer: (C) Phosphorus

   
Answer Explanation:
        
The binding energy decreases as you move to outer shells. The 3s subshell electrons are less tightly bound than inner electrons. The peak at 6.5 MJ/mol corresponds to the 3s subshell.
        
Correct Answer: (B) The peak at 6.5 MJ/mol

  
Answer Explanation:
        
The electron with the highest binding energy (largest Coulombic attraction to the nucleus) is the one in the 1s subshell. The peak at 103 MJ/mol represents the 1s electrons.
        
Correct Answer: (A) The peak at 103 MJ/mol

    
Answer Explanation:
        
The valence electrons are those in the outermost energy level. For phosphorus (\( \text{P}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^3 \)), there are 5 valence electrons (3s2 and 3p3).
        
Correct Answer: (C) 5

  
Answer Explanation:

        
Atomic radius decreases across a period from left to right due to increasing nuclear charge attracting electrons more strongly.

        
All given elements are in the second period of the periodic table. As we move from Lithium (Li) to Neon (Ne), the atomic number increases, and the atomic radius decreases.

        
Therefore, Neon (Ne) has the smallest atomic radius among the options.

        
Correct Answer: (D) Neon (Ne)

    
Answer Explanation:
        
First ionization energy increases across a period and decreases down a group. Argon is a noble gas with a full valence shell, so it has the highest first ionization energy among the options.
        
Correct Answer: (D) Argon (Ar)

   
Answer Explanation:
        
Lattice energy increases with the charge of the ions and decreases with the size of the ions. Aluminum ion has the highest positive charge (+3), leading to the greatest lattice energy when combined with chloride.
        
Correct Answer: (C) Aluminum ion (\( \text{Al}^{3+} \))

 
Answer Explanation:

        
The general trend in first ionization energy increases across a period due to increasing nuclear charge attracting the electrons more strongly. However, deviations occur between groups 2 and 13 (Be and B) and between groups 15 and 16 (N and O).

        
Between Beryllium (Be) and Boron (B):

        
Beryllium has the electron configuration 1s² 2s², with a fully filled 2s subshell. Boron has the electron configuration 1s² 2s² 2p¹. The 2p electron in Boron is at a higher energy level than the 2s electrons in Beryllium and is shielded by the 2s electrons, making it easier to remove despite the higher nuclear charge. Therefore, Boron has a lower first ionization energy than Beryllium.

        
Between Nitrogen (N) and Oxygen (O):

        
Nitrogen has the electron configuration 1s² 2s² 2p³, with a half-filled p subshell, which is relatively stable. Oxygen has the electron configuration 1s² 2s² 2p⁴, where one of the p orbitals contains a pair of electrons. The electron-electron repulsion in this doubly occupied p orbital makes it easier to remove an electron from Oxygen than from Nitrogen. Thus, Oxygen has a lower first ionization energy than Nitrogen.

        
Option (B) correctly explains the deviation observed between Nitrogen and Oxygen due to increased electron-electron repulsion in Oxygen's p orbital.

        
Correct Answer: (B) Oxygen has a lower first ionization energy than Nitrogen due to increased electron-electron repulsion in the doubly occupied p orbital.

 
Answer Explanation:
        
The 2s orbital penetrates closer to the nucleus due to its radial nodes, leading to lower energy because of increased nuclear attraction.
        
Correct Answer: (B) The 2s orbital penetrates closer to the nucleus.

 
Answer Explanation:
        
The 5p5 subshell has one unpaired electron because p orbitals can hold 6 electrons, and having 5 electrons means one orbital is unpaired.
        
Correct Answer: (B) 1

    
Answer Explanation:
        
Atomic radius increases down a group and decreases across a period. Among the given elements, Carbon (C) is furthest to the left in the same period, so it has the largest atomic radius.
        
Correct Answer: (D) Carbon (C)

   
Answer Explanation:

        
The expected electron configuration of Chromium (Cr) based on the Aufbau principle is [Ar] 4s² 3d⁴. However, Chromium actually has the electron configuration [Ar] 4s¹ 3d⁵.

        
This anomaly occurs because a half-filled d subshell (3d⁵) provides extra stability. By promoting one electron from the 4s orbital to the 3d orbital, Chromium achieves a half-filled d subshell and a half-filled s subshell, which lowers the energy of the atom.

        
The other elements listed do not exhibit this anomaly:

        

            
Vanadium (V): [Ar] 4s² 3d³
            
Manganese (Mn): [Ar] 4s² 3d⁵
            
Iron (Fe): [Ar] 4s² 3d⁶
        

        
Correct Answer: (B) Chromium (Cr)

   
Answer Explanation:
        
Oxygen is smaller and thus experiences greater electron-electron repulsion when gaining an electron, making its electron affinity more negative compared to sulfur. Sulfur, being larger, has a less negative electron affinity because the added electron experiences less repulsion.
        
Correct Answer: (B) Sulfur has a larger atomic radius, reducing electron-electron repulsion.

    
Answer Explanation:
        
The large jump in ionization energy after the third ionization (from 3,650 kJ/mol to 25,000 kJ/mol) suggests that the element has three valence electrons. Therefore, the element likely has 3 valence electrons.
        
Correct Answer: (C) 3

   
Answer Explanation:
        
Atomic radius increases down a group and decreases across a period. Potassium is further down and to the left of the periodic table than the other options, giving it the largest atomic radius.
        
Correct Answer: (A) Potassium (K)

Answer Explanation:
 
The 3d subshell has five orbitals, each capable of holding two electrons. With 9 electrons in the 3d subshell, four orbitals are fully filled, and the fifth orbital has one electron, which is unpaired. Therefore, there is 1 unpaired electron in the 3d subshell.
 
Correct Answer: (A) 1

Answer Explanation:
 
All the listed ions achieve a total of 10 electrons, matching the electron configuration of neon (Ne):
 
 
\( \text{Na}^+ \): Sodium loses one electron (11 - 1 = 10 electrons).
 
\( \text{O}^{2-} \): Oxygen gains two electrons (8 + 2 = 10 electrons).
 
\( \text{F}^- \): Fluorine gains one electron (9 + 1 = 10 electrons).
 
 
Therefore, all of the above ions have the same electron configuration as neon.
 
Correct Answer: (D) All of the above

  
Answer Explanation:

        
To identify element X, we need to analyze the successive ionization energies and look for significant jumps that indicate the removal of core electrons.

        
From the data provided:

        

            
IE₁ to IE₂: 578 to 1817 kJ/mol (increase of 1239 kJ/mol)
            
IE₂ to IE₃: 1817 to 2745 kJ/mol (increase of 928 kJ/mol)
            
IE₃ to IE₄: 2745 to 11577 kJ/mol (increase of 8832 kJ/mol)
            
IE₄ to IE₅: 11577 to 14831 kJ/mol (increase of 3254 kJ/mol)
        

        
The significant jump occurs between the third and fourth ionization energies (IE₃ and IE₄). This suggests that after removing three electrons, we begin to remove core electrons from a lower energy level.

        
This indicates that element X has three valence electrons.

        
Now, let's consider the options:

        

            
Magnesium (Mg): Has 2 valence electrons (electron configuration [Ne] 3s²)
            
Aluminum (Al): Has 3 valence electrons (electron configuration [Ne] 3s² 3p¹)
            
Silicon (Si): Has 4 valence electrons (electron configuration [Ne] 3s² 3p²)
            
Phosphorus (P): Has 5 valence electrons (electron configuration [Ne] 3s² 3p³)
        

        
Therefore, element X is most likely Aluminum (Al), which has three valence electrons, consistent with the significant increase after the third ionization energy.

        
Correct Answer: (B) Aluminum (Al)

Answer Explanation:

The s-orbitals are spherical in shape. This means the probability of finding an electron is the same at any point equidistant from the nucleus.

Correct Answer: (A) s-orbital

Answer Explanation:
Beryllium has a completely filled 2s subshell (2s²), which provides extra stability and requires more energy to remove an electron. In contrast, boron has an electron configuration of 2s² 2p¹, where the single 2p electron is easier to remove compared to the filled 2s electrons in beryllium. This results in beryllium having a higher first ionization energy than boron.
Correct Answer: (C) Beryllium has a completely filled s subshell, resulting in higher ionization energy.
       

   
Answer Explanation:
        
Nitrogen has three unpaired electrons in its p orbitals (1s2 2s2 2p3). Carbon has two, oxygen has two, and fluorine has one unpaired electron in their p orbitals.
        
Correct Answer: (C) Nitrogen (N)

Answer Explanation:
        
The 4p3 subshell has three unpaired electrons because p orbitals can hold a maximum of 6 electrons, and having 3 electrons means each p orbital contains one unpaired electron.
        
Correct Answer: (D) 3

 
Answer Explanation:
        
Nitrogen has a half-filled 2p subshell (2p3), which is more stable. Removing an electron from oxygen results in a half-filled subshell, so it's slightly easier to ionize than nitrogen.
        
Correct Answer: (B) Nitrogen has a half-filled p subshell, making it more stable.

 
Answer Explanation:
        
Fluorine is the most electronegative element on the periodic table.
        
Correct Answer: (B) Fluorine (F)

 
Answer Explanation:

        
The energy (\( E \)) of a photon is related to its frequency (\( \nu \)) by the equation:

        
\( E = h\nu \), where \( h \) is Planck's constant.

        
Frequency and wavelength (\( \lambda \)) are related by the speed of light (\( c \)) through the equation:

        
\( c = \lambda\nu \). This means that as the wavelength increases, the frequency decreases, and vice versa.

        
From these relationships, we can conclude that the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength.

        
Correct Answer: (C) The energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength.

    
Answer Explanation:
        
First ionization energy generally increases across a period and decreases down a group. Additionally, elements with nearly full valence shells or those that have half-filled p subshells tend to have higher ionization energies due to increased stability.
        
Among the given elements:
        

            
Silicon (Si): Positioned further left in the period, thus having a relatively lower ionization energy.
            
Phosphorus (P): Has a half-filled 3p subshell, which adds extra stability, resulting in a relatively high ionization energy.
            
Sulfur (S): Although further right than phosphorus, the added electron pairing in the 3p subshell causes slight electron-electron repulsion, slightly reducing its ionization energy compared to phosphorus.
            
Fluorine (F): Located in the second period and near the noble gas configuration, it has a very high first ionization energy due to its small atomic radius and high effective nuclear charge.
        
        
Therefore, fluorine (F) has the highest first ionization energy among these elements due to its position near the top right of the periodic table.
        
Correct Answer: (D) Fluorine (F)

  
Answer Explanation:

        
To identify the correct metal, we first consider the neutral atom's electron configuration and how it changes upon ion formation.

        
(A) Iron (Fe): The neutral iron atom has the electron configuration [Ar] 4s2 3d6. When it forms a 3+ ion (Fe3+), it loses two electrons from the 4s orbital and one electron from the 3d orbital, resulting in the configuration [Ar] 3d5.

        
(B) Cobalt (Co): The neutral cobalt atom has the electron configuration [Ar] 4s2 3d7. When it forms a 3+ ion (Co3+), it loses two electrons from the 4s orbital and one electron from the 3d orbital, resulting in the configuration [Ar] 3d6.

        
(C) Nickel (Ni): The neutral nickel atom has the electron configuration [Ar] 4s2 3d8. When it forms a 3+ ion (Ni3+), it loses two electrons from the 4s orbital and one electron from the 3d orbital, resulting in the configuration [Ar] 3d7.

        
(D) Copper (Cu): The neutral copper atom has the electron configuration [Ar] 4s1 3d10. When it forms a 3+ ion (Cu3+), it loses one electron from the 4s orbital and two electrons from the 3d orbital, resulting in the configuration [Ar] 3d8.

        
Thus, the transition metal that has the electron configuration [Ar] 3d6 when it forms a 3+ ion is:

        
Correct Answer: (B) Cobalt (Co)

   
Answer Explanation:
        
Sodium has the highest second ionization energy because its first electron is removed from the 3s subshell, but its second electron is removed from the stable noble gas core (2p), requiring significantly more energy.
        
Correct Answer: (A) Sodium (Na)

  
Answer Explanation:

        
In a period, ionization energy generally increases due to increasing nuclear charge. However, Oxygen has a lower first ionization energy than Nitrogen. This is because, in Oxygen, one of the p orbitals is doubly occupied, resulting in increased electron-electron repulsion, which makes it easier to remove an electron.

        
Regarding electronegativity, Fluorine is the most electronegative element in the periodic table, with an electronegativity higher than both Oxygen and Nitrogen.

        
Correct Answer: (A) Oxygen has a lower first ionization energy than Nitrogen, and its electronegativity is lower than that of Fluorine.

  
Answer Explanation:
        
The large jump in ionization energy after the second ionization suggests that the element has two valence electrons. Therefore, the element likely has 2 valence electrons.
        
Correct Answer: (B) 2
  

 
Answer Explanation:
        
Ionization energy decreases down a group and increases across a period. Lithium is the furthest to the left in its period, so it has the smallest first ionization energy among the options.
        
Correct Answer: (D) Lithium (Li)
 

  
Answer Explanation:
        
Metallic character increases down a group and decreases across a period. Aluminum is the furthest to the left among the given elements and is a metal, giving it the greatest metallic character.
      

Correct Answer: (A) Aluminum (Al)

     
Answer Explanation:
        
The lanthanide series involves the filling of the 4f subshell.
        
Correct Answer: (A) 4f
   

     
Answer Explanation:
        
The element has electrons ending in 4s2, placing it in Period 4, Group 2A of the periodic table. The element that fits this description is calcium.
        
Correct Answer: (A) Calcium (Ca)
 

Answer Explanation:
 
Chromium (Cr) exhibits an anomalous electron configuration. Instead of the expected [Ar] 4s2 3d4, chromium's actual electron configuration is [Ar] 4s1 3d5. This half-filled d subshell provides extra stability due to the symmetrical distribution of electrons.
 
Correct Answer: (C) Chromium (Cr)

Answer Explanation:
 
Magnesium (Mg) has two valence electrons in the 3s subshell. The first and second ionization energies involve removing these valence electrons. The third ionization energy requires removing an electron from the filled 2p subshell (a core electron), resulting in a significant increase in ionization energy.
 
Correct Answer: (A) Magnesium (Mg)

       
Answer Explanation:
        
Fluorine has the highest electron affinity among the given elements because it is highly electronegative and readily accepts an electron to complete its valence shell.
        
Correct Answer: (A) Neon (Ne)
 

      
Answer Explanation:

        
Oxygen (O) has a higher electronegativity than Nitrogen and also exhibits an exception in the ionization energy trend. Oxygen has a lower first ionization energy than Nitrogen due to increased electron-electron repulsion in the doubly occupied p orbital, which makes it easier to remove an electron.

        
Correct Answer: (B) Oxygen (O), which has a higher electronegativity and a lower first ionization energy than Nitrogen due to electron-electron repulsion in the p orbital.
 

      
Answer Explanation:
        
In a period, atomic size decreases from left to right due to increasing effective nuclear charge. Therefore, element X, being the smallest, has the greatest effective nuclear charge and the highest first ionization energy among the options. Electron shielding is similar across a period, and electronegativity increases across a period, making it greatest in element X, not Z.
        
Correct Answer: (B) The first ionization energy will be greatest in element X.
  

Answer Explanation:

For transition metals, each successive electron removal requires more energy, so the ionization energy increases.

Correct Answer: (B)

       
Answer Explanation:
        
The number of peaks in a photoelectron spectrum corresponds to the number of different electron shells or subshells being ionized. An atom with 10 protons is neon (Ne), which has the electron configuration 1s2 2s2 2p6. This results in three distinct binding energy peaks corresponding to the 1s, 2s, and 2p subshells.
        
Correct Answer: (B) Atom with 10 protons and 10 neutrons
  

    
Answer Explanation:
        
The large jump between the second and third ionization energies indicates that the element has two valence electrons. After removing two electrons, the third ionization energy is much higher because it requires removing an electron from a stable, lower-energy inner shell. Magnesium (Mg) has two valence electrons, which aligns with the provided ionization energy values.
        
Correct Answer: (B) Magnesium
  

    
Answer Explanation:
        
When phosphorus gains three electrons to form \( \text{P}^{3-} \), the increased number of electrons leads to greater electron-electron repulsion in the valence shell. This repulsion causes the ion to expand, resulting in a larger radius compared to the neutral atom. The nuclear charge remains unchanged, but the added electrons increase the repulsive forces among electrons.
        
Correct Answer: (D) The electrons in \( \text{P}^{3-} \) have greater Coulombic repulsion than those in the neutral atom.
  

  
Answer Explanation:
        
First, calculate the moles of each element:
        
Moles of Si: \( \dfrac{14.0\, \text{g}}{28.09\, \text{g/mol}} \approx 0.498 \, \text{mol} \)
        
Moles of O: \( \dfrac{32.0\, \text{g}}{16.00\, \text{g/mol}} = 2.00 \, \text{mol} \)
        
Next, divide by the smallest number of moles to find the simplest ratio:
        
Si: \( \dfrac{0.498}{0.498} = 1 \)
        
O: \( \dfrac{2.00}{0.498} \approx 4 \)
        
The empirical formula is SiO2.
        
Correct Answer: (a) SiO4
 

      
Answer Explanation:

        
Lithium (Li) has a low electronegativity and a single valence electron in its 2s orbital. Removing this first electron requires relatively little energy. However, removing the second electron requires breaking into the stable, noble gas-like 1s core, resulting in a substantial increase in ionization energy.

        
Correct Answer: (A) Lithium (Li), which has a low electronegativity and shows a massive jump between its first and second ionization energies.
  

      
Answer Explanation:
        
Within the same period, nonmetals have higher ionization energies compared to metals. This is because nonmetals are closer to achieving a filled valence shell, making it energetically more difficult to remove an electron. Additionally, nonmetals typically have a higher effective nuclear charge and smaller atomic radii, which further increase their ionization energies.
        
Correct Answer: (D) Nonmetals have higher ionization energies because they are closer to having filled a complete energy level.
  

      
Answer Explanation:
        
Electronegativity generally increases across a period from left to right. Nitrogen is more electronegative than phosphorus because it is smaller and has a higher effective nuclear charge, which attracts electrons more strongly. However, oxygen is more electronegative than nitrogen due to having a higher nuclear charge and a smaller atomic radius, enhancing its ability to attract electrons.
        
Correct Answer: (A) Nitrogen's value is greater than that of phosphorus because nitrogen is smaller, but less than that of oxygen because nitrogen has a smaller effective nuclear charge.
 

    
Answer Explanation:
        
Paramagnetic species have unpaired electrons. Let's examine each ion:
        

            
(A) \( \text{Mn}^{2+} \): Electron configuration [Ar] 3d5, all five d electrons are unpaired.
            
(B) \( \text{Ni}^{2+} \): Electron configuration [Ar] 3d8, two unpaired electrons.
            
(C) \( \text{Ti}^{2+} \): Electron configuration [Ar] 3d2, two unpaired electrons.
            
(D) \( \text{Cr}^{2+} \): Electron configuration [Ar] 3d4, four unpaired electrons.
        
        
However, chromium often exhibits a stable half-filled configuration, but for \( \text{Cr}^{2+} \), the configuration is [Ar] 3d4, which has four unpaired electrons. Manganese ion \( \text{Mn}^{2+} \) has five unpaired electrons, which is the most among the options.
        
Correct Answer: (A) \( \text{Mn}^{2+} \)

    
Answer Explanation:
        
Transition metals commonly exhibit a +2 oxidation state because they lose their outermost s-electrons first when forming cations. For example, elements like iron (Fe) lose the two 4s electrons to form Fe2+.
        
Correct Answer: (D) Transition metals will lose their outermost s-block electrons when forming bonds.